{"id":1405,"date":"2015-10-01T12:41:58","date_gmt":"2015-10-01T12:41:58","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1405"},"modified":"2024-08-23T12:55:56","modified_gmt":"2024-08-23T12:55:56","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-colisoes-mecanicas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/colisoes-mecanicas-ou-choques-mecanicos\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-colisoes-mecanicas\/","title":{"rendered":"Colis\u00f5es Mec\u00e2nicas – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Colis\u00f5es Mec\u00e2nicas<\/b><\/span><\/span><\/span><\/p>\n

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

a) 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=(10 + V\u2019<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>)\/(5 + 10)\u00a0 —\u00a0 15=10 +V\u2019<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 V\u2019<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=5m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) e=V<\/b><\/span><\/span><\/span>rdepois<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>rantes<\/b><\/span><\/span><\/sub><\/span>=(4 \u2013 2)\/(5 \u2013 1)\u00a0 —\u00a0\u00a0e=0,5 (choque parcialmente el\u00e1stico)<\/b><\/span><\/span><\/span><\/p>\n

c) 0.8=V\u2019<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\/15\u00a0\u00a0—\u00a0 V\u2019<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>=8m\/s<\/b><\/span><\/span><\/span><\/p>\n

d) 0,6=(V\u2019<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 40)\/(40 + 60)\u00a0 —\u00a0 60=V\u2019<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 40\u00a0 —\u00a0\u00a0V\u2019<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>=100 km\/h<\/b><\/span><\/span><\/span><\/p>\n

e) e=0\/4\u00a0 —\u00a0\u00a0e=0 (choque inel\u00e1stico)<\/b><\/span><\/span><\/span><\/p>\n

f) e=\u221a4\/16\u00a0\u00a0—\u00a0 e=0,25 (choque parcialmente el\u00e1stico)<\/b><\/span><\/span><\/span><\/p>\n

02<\/b><\/span><\/span><\/span>–\u00a0Considere o choque perfeitamente el\u00e1stico.<\/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\u00a0V\u2019<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=m\u00a0 —\u00a0\u00a0 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>=mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)V\u2019=2mV\u2019 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=2mV\u2019\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 mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2mV\u2019\u00a0 —\u00a0\u00a0V\u2019=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2 \u2013 R- D<\/b><\/span><\/span><\/span><\/p>\n

04<\/b><\/span><\/span><\/span>–\u00a0Considerando as velocidades para a direita como positivas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Antes\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=m30 + m.20=50.m —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=50.m\u00a0 —\u00a0 depois\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=mV\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>+ mV\u2019=2mV\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 50m=2mV\u2019\u00a0 —\u00a0\u00a0V\u2019=25m\/s<\/b><\/span><\/span><\/span><\/p>\n

05-\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\u00a0V\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\u0394E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=18 \u2013 12 = 6j<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0Considerando velocidade4s positivas para a direita\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=5.1 + 1.(-8)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>= -3kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(5 + 1)V\u2019\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=6V\u2019\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/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 – 3 =6V\u2019\u00a0 —\u00a0 V\u2019= – 0,5m\/s (negativa, para a esquerda) R- A<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0Estabelecendo velocidades para a esquerda como positivas\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=(30 + 10).0 + 20.V<\/b><\/span><\/span><\/span>Z<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=20V<\/b><\/span><\/span><\/span>Z<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(30 + 10 +20).0,5\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=30kg.m\/s\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 20V<\/b><\/span><\/span><\/span>z<\/b><\/span><\/span><\/sub><\/span>=30\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>z<\/b><\/span><\/span><\/sub><\/span>=1,5m\/s<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0Trata-se de um choque inel\u00e1stico (e=0) e eles, ap\u00f3s o choque, se movem juntos com velocidade comum de 8m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>=2.12 + m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>.4\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=24 + 4.m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(m<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>).V\u2019=(2 + m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>).8\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=16 + 8m<\/b><\/span><\/span><\/span>Y<\/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 24 + 4m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>=16 + 8m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 8=4m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0m<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>=2kg<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=30V\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(30 + m\u2019).V\/2\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 30V=(30 + m\u2019).V\/2 \u00a0\u00a0—\u00a0 60=30 + m\u2019\u00a0 —\u00a0m\u00b4=30g R- D —\u00a0\u00a0ou, o 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 bolinha de isopor. R- D<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0a) Considerando velocidades para a esquerda como positivas\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=6.000kg \u2013 V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=54km\/h\/3,6=15m\/s\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=2.000kg \u2013<\/b><\/span><\/span><\/span><\/p>\n

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

V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=72km\/h\/3,6=20m\/s\u00a0 —\u00a0 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>c<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=2.000.20 + 6.000.15\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=130.000kg.m\/s\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>a<\/b><\/span><\/span><\/sub><\/span>)V\u2019=8.000V\u2019\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=8.000V\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 130.000=8.000V\u2019\u00a0 —\u00a0 V\u2019=16,25m\/s X 3,6\u00a0\u00a0 —\u00a0\u00a0V\u2019=58,5 km\/h<\/b><\/span><\/span><\/span><\/p>\n

b) Energia cin\u00e9tica inicial do carro \u2013 E<\/b><\/span><\/span><\/span>ci<\/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>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.000.(20)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=400.000J\u00a0 —\u00a0 Energia potencial gravitacional de centro de massa do carro que subiu h=50cm=0,5m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.g.h=2.000.10.0,5\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=10.000J\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 400.000J \u2013 100%\u00a0 —\u00a0 10.000J \u2013 p\u00a0 —\u00a0\u00a0 p=10\/4\u00a0\u00a0—\u00a0 p=2,5%<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0Q<\/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 —\u00a0 80.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.8\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=80.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/8\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.1\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>kg=2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/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>=10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.8 + 2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.1\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)V\u2019=3,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.V\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 10,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=3,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.V\u2019\u00a0 —\u00a0\u00a0V\u2019=3m\/s\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

12-\u00a0<\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0E<\/b><\/span><\/span><\/span><\/span>p<\/b><\/span><\/span><\/span><\/sub><\/span>=Mgh=(0,01 + 0,190).10.0,1\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span><\/span>p<\/b><\/span><\/span><\/span><\/sub><\/span>=0,2.10.0,1\u00a0 —\u00a0\u00a0E<\/b><\/span><\/span><\/span><\/span>p<\/b><\/span><\/span><\/span><\/sub><\/span>=0,2J<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) C\u00e1lculo da velocidade (V<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>) com que o conjunto bloco esfera de massa M=0,2kg sai de Q e chega em P com velocidade nula\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cP<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pP<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cQ<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pQ<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0 + 0,2=MV<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0\u00a0 —\u00a0 0,2=0,2V<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=\u221a2m\/s\u00a0 —\u00a0 choque inel\u00e1stico da esfera de massa m= 0,01kg que chega com velocidade V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0com o bloco de massa m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>=0,190kg que est\u00e1 em repouso sendo que, ap\u00f3s o choque o conjunto de massa M=0,2kg se move com V\u2019=\u221a2m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=0,01.V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=0,01V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=0,2V\u2019=0,2.\u221a2\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 0,01V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=0,2.\u221a2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=20.\u221a2\u224820.1,4\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u224828m\/s<\/b><\/span><\/span><\/span><\/p>\n

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

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—\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\u00a0\u00a0—\u00a0\u00a0M<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=100kg\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-\u00a0a)\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 6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 0 = (3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 1,8.10<\/b><\/span><\/span><\/span>27<\/b><\/span><\/span><\/sup><\/span>)V\u2019\u00a0 —\u00a0 observe que 3.10+14kg<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00e9 praticamente desprez\u00edvel em rela\u00e7\u00e3o a 1,8.10<\/b><\/span><\/span><\/span>27<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 18.10<\/b><\/span><\/span><\/span>18<\/b><\/span><\/span><\/sup><\/span>=1,8.10<\/b><\/span><\/span><\/span>27<\/b><\/span><\/span><\/sup><\/span>V\u2019\u00a0 —\u00a0 V\u2019=10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>m\/s (esse valor \u00e9 desprez\u00edvel e esse choque n\u00e3o afetou a velocidade de J\u00fapiter)<\/b><\/span><\/span><\/span><\/p>\n

b) A energia mec\u00e2nica total dissipada foi toda energia cin\u00e9tica perdida pelo cometa no choque\u00a0 — E<\/b><\/span><\/span><\/span>dissipada<\/b><\/span><\/span><\/sub><\/span>=MV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>.(6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>dissipada<\/b><\/span><\/span><\/sub><\/span>=5,4.10<\/b><\/span><\/span><\/span>23<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0e=\u221a6\/12\u00a0 —\u00a0 e=\u221a2\/2<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0m<\/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 —\u00a0 aplicando<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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\u00a0a declara\u00e7\u00e3o \u00e9 falsa, pois a velocidade do carro era aproximadamente de 145km\/h<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\/m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=1\/999\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=999m<\/b><\/span><\/span><\/span>p<\/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 m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>+ 0=(m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>).0,25\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=(m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 999m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>).0,25\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=1.000m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>.0,25\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=250m\/s R- C<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0Trata-se de um choque inel\u00e1stico obl\u00edquo (se movem juntos ap\u00f3s o choque)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

19-<\/b><\/span><\/span><\/span>\u00a0Esquematizando a situa\u00e7\u00e3o e supondo que ap\u00f3s o choque, eles se movam para a direita<\/b><\/span><\/span><\/span><\/p>\n

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

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\u00a013.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\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\u00a0V<\/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\u00a0resolvendo o sistema composto por I e II\u00a0 —\u00a0\u00a0V<\/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

20-<\/b><\/span><\/span><\/span>\u00a0Esquematizando a situa\u00e7\u00e3o<\/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 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.6 + 0 = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.4 + m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.6\u00a0 —\u00a0 2m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=6m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3\u00a0 —\u00a0 e=V<\/b><\/span><\/span><\/span>rd<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>ra<\/b><\/span><\/span><\/sub><\/span>=(6-4)\/6\u00a0 —\u00a0\u00a0e=1\/3 R- C<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0a) 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\u00a0F<\/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\u00a0e=3\/4<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0Q<\/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>1<\/b><\/span><\/span><\/sub><\/span>.(-2) + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.(4) = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.(3) + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.(1)\u00a0\u00a0—\u00a0 5m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Como n\u00e3o \u00e9 inel\u00e1stico (n\u00e3o se movem juntos) nem el\u00e1stico (enunciado), \u00e9 parcialmente el\u00e1stico\u00a0 —\u00a0 antes\u00a0 —Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=MV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 M2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 depois\u00a0 —\u00a0 como B p\u00e1ra, A s\u00f3 pode voltar com velocidade V\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=M.(-V)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=-MV\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 MV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 2MV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

-MV\u00a0 —\u00a0\u00a0V=-V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=MV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=5MV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 X 2\/MV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/5\u00a0\u00a0 —\u00a0 e=m\u00f3dulo da velocidade relativa depois \/ m\u00f3dulo da velocidade relativa antes\u00a0 —\u00a0 e=V<\/b><\/span><\/span><\/span>rd<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>ra<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0e=1\/3\u00a0\u00a0\u00a0R-D<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0O 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\u00a0V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=Vo\/3\u00a0\u00a0V<\/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\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 (2pR)\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>) + (2pR) =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\u03c0R=3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 t=\u03c0R\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0I\u00a0 —\u00a0 substituindo I em\u00a0 S<\/b><\/span><\/span><\/span>A\u00a0<\/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>.\u03c0R\/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>=\u03c0R (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>.\u03c0R\/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\u03c0R (percorreu uma circunfer\u00eancia e meia a partir do ponto 1) \u2013\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0Esquematizando a situa\u00e7\u00e3o antes e depois do choque da bola de t\u00eanis com o caminh\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/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 0,6=(10 + V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019)\/(20 \u2013 10)\u00a0 —\u00a0 6=10 + V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019= -4m\/s ( a bola retorna com velocidade de 4m\/s)<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0Como 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>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0Como o choque \u00e9 perfeitamente el\u00e1stico e elas tem a mesma massa, ap\u00f3s o choque, trocam suas velocidades\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0 e V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0Colis\u00e3o perfeitamente el\u00e1stica de A com B e eles n\u00e3o trocam suas velocidades, pois suas massas<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

s\u00e3o diferentes 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\u00a0V<\/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>+ 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\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>(01) Falsa \u2013 somente se o choque for perfeitamente el\u00e1stico<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/sub><\/span>(02) m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0 = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 + m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0 —\u00a0 2V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= 2V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 + V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019 I\u00a0 —\u00a0 1=(V\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019)\/V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 — V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 = V<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>II\u00a0 —\u00a0 I com II\u00a0 — V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/3 e V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019=4V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/3 \u2013 Falsa<\/b><\/span><\/span><\/span><\/p>\n

(4) Verdadeira \u2013 choque perfeitamente el\u00e1stico e mesma massa, trocam suas velocidades<\/b><\/span><\/span><\/span><\/p>\n

(08) Falsa \u2013 a quantidade de movimento sempre \u00e9 a mesma antes e depois do choque<\/b><\/span><\/span><\/span><\/p>\n

(16) m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0=m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 + 2m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0 —\u00a0\u00a0 ) 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>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 + 2m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0 — V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 + 2V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019 I\u00a0 —\u00a0\u00a0 1=(V\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019)\/V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 — II\u00a0 — V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019 \u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 resolvendo I com II\u00a0 — V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019=-V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/3 (sentidos opostos) \u2013 Verdadeiro<\/b><\/span><\/span><\/span><\/p>\n

\u00a0As corretas s\u00e3o 4 e 16.<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0A 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

31-<\/b><\/span><\/span><\/span>\u00a0Velocidade com que o bloco de massa m chega \u00e0 base da rampa (pontoC) —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh = mV<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0 + m.10.0,8 = mV<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0\u00a0 —\u00a0 8=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=4m\/s\u00a0 —\u00a0 como n\u00e3o existe atrito, ele colide elasticamente em B com o outro bloco de mesma massa e eles trocam suas velocidades \u2013\u00a0R- 4m\/s<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0a) Primeira colis\u00e3o em\u00a0\u00a0P\u00a0\u2013 A percorre 1,6m e se choca com B \u2013 como n\u00e3o existe atrito, o movimento \u00e9 com velocidade constante de + 2m\/s (para a direita) e trata-se de um MRU de equa\u00e7\u00e3o V=\u0394S\/\u0394t\u00a0 —\u00a0 2=1,6\/\u0394t\u00a0 —\u00a0 \u0394t=0,8s (t<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=0,8s).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Sendo o choque perfeitamente el\u00e1stico, em P, A e B trocam suas velocidades, A p\u00e1ra e B se move com velocidade de +2m\/s\u00a0 at\u00e9 chegar em Q, base da rampa, e, nesse percurso demora tamb\u00e9m 0,8s (t<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=1,6s).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em seguida, sobe a rampa at\u00e9 chegar \u00e0 altura m\u00e1xima H, (ponto R), com velocidade zero no instante t<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=2,0s (dado do exerc\u00edcio).<\/b><\/span><\/span><\/span><\/p>\n

Observe que nessa subida demorou 0,4s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A partir de R, come\u00e7a a descer a rampa, demorando tamb\u00e9m 0,4s, chegando em Q, no instante t\u2019<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=2,4s, com velocidade de<\/b><\/span><\/span><\/span><\/p>\n

-2,0m\/s<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Retorna \u00e0 P, demorando 0,8s, aonde chega no instante t\u2019<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=3,2s<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em P, colide com A, trocando suas velocidades e A chegando \u00e0 parede, em t=4,0s (3,2s + 0,8s), onde recome\u00e7a tudo novamente.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A seguir, temos um esquema do movimento durante a ida e a volta:<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

c) T = 4,0s.<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0O 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 aC, 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

34-<\/b><\/span><\/span><\/span>\u00a0Soma vetorial das quantidades de movimento antes dos choques:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Baseado no Princ\u00edpio da Conserva\u00e7\u00e3o da quantidade de Movimento (\"\"), dentre todas as alternativas, a \u00fanica cuja soma vetorial das quantidades de movimento depois do choque tamb\u00e9m \u00e9 vertical e para cima \u00e9 a\u00a0e\u00a0\u00a0\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0A for\u00e7a de atrito (F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>) contr\u00e1ria ao movimento \u00e9 a \u00fanica\u00a0 for\u00e7a, na dire\u00e7\u00e3o do movimento,que age sobre P diminuindo-lhe a velocidade\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0 Fr=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=10N\u00a0 —\u00a0 10=5.a\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Aplicando Torricelli com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0 —\u00a0 a= -2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(retardando)\u00a0 —\u00a0 \u0394S=12m\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>\u00a0+ 2.a.\u0394S\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-2).12\u00a0 —\u00a0 V=\u221a52\u00a0 —\u00a0 sendo o choque perfeitamente el\u00e1stico e os corpos tendo a mesma massa, ap\u00f3s o choque eles trocam suas velocidades, P p\u00e1ra e Q segue com velocidade constante de 7,2m\/s at\u00e9 chegar \u00e0 base da rampa (ponto R). A partir da\u00ed sua velocidade come\u00e7a a diminuir at\u00e9 chegar ao ponto C, onde se anula.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Princ\u00edpio da Conserva\u00e7\u00e3o da Energia Mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= mV<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h\u00a0 —\u00a0 5.(\u221a52)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0 = 0 + 5.10.h\u00a0 —\u00a0 26 = 10h \u00a0—\u00a0 h=2,6m\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

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

36-\u00a0<\/b><\/span><\/span><\/span>O sistema \u00e9 mecanicamente isolado e conservativo\u00a0 —\u00a0 assim, para determinar as velocidades dos discos depois do choque (\"\"\u00a0e\u00a0\"\") voc\u00ea pode \u00a0usar a conserva\u00e7\u00e3o da quantidade de movimento e a conserva\u00e7\u00e3o da energia mec\u00e2nica,pois voc\u00ea tem duas equa\u00e7\u00f5es com duas inc\u00f3gnitas (\"\"\u00a0e\u00a0\"\") que s\u00e3o\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.V<\/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>\u2019 + M<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019 e\u00a0 M<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= (M<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\/2 + (M<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\/2\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

OBS: seria bem mais f\u00e1cil usar a conserva\u00e7\u00e3o da quantidade de movimento e o coeficiente de restitui\u00e7\u00e3o (e = 1), uma vez que o choque \u00e9 perfeitamente el\u00e1stico.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

37-\u00a0As duas bolas t\u00eam mesma massa (m)\u00a0 —\u00a0 desprezando a resist\u00eancia do ar, se elas s\u00e3o largadas da mesma altura, chegar\u00e3o ao solo com mesma velocidade (v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 orientando a trajet\u00f3ria para cima, como mostrado a seguir, e aplicando o teorema do impulso nos dois casos\u00a0 —\u00a0\"\"\u00a0\u00a0—\u00a0 \u00a0\u00a0Ib = m |v + v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>|\u00a0 —\u00a0 Im = m |0 \u2013 (\u2013v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)|\u00a0 —\u00a0 Im = m |v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>|\u00a0 —\u00a0\u00a0Ib > Im.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0Dados\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 800 kg\u00a0 —\u00a0\u00a0 v<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 90 km\/h = 25 m\/s\u00a0 —\u00a0\u00a0 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 450 kg e v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 120 km\/h =120\/3,6=100\/3 m\/s\u00a0 —\u00a0 lembre-se de que voc\u00ea n\u00e3o deve fazer uma divis\u00e3o que d\u00e1 d\u00edzima no meio da solu\u00e7\u00e3o de um exerc\u00edcio\u00a0 —\u00a0 trabalhe com a fra\u00e7\u00e3o\u00a0 —\u00a0 se na resposta final a d\u00edzima persistir, a\u00ed sim, fazem-se as contas e os arredondamentos\u00a0 — note-se que se fosse feita a divis\u00e3o nessa quest\u00e3o, obtendo 33,3 m\/s para v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, ter\u00edamos um tremendo trabalho e n\u00e3o chegar\u00edamos a resposta exata.\u00a0 —\u00a0 m\u00f3dulo da quantidade de movimento dos dois carros antes da colis\u00e3o\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>1<\/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>=800.25\u00a0 — Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=20.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=450.100\/3\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=15.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg.m\/s —\u00a0 como quantidade de movimento \u00e9 uma grandeza vetorial, como mostra o esquema, vem\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(20.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (15.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg.m\/s\u00a0 —\u00a0Sendo a colis\u00e3o inel\u00e1stica eles se movem juntos ap\u00f3s a mesma com velocidade V e massa total\u00a0 —\u00a0\u00a0 M=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 M=1.250kg\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>=MV\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>=1250V\u00a0 —\u00a0 como o sistema \u00e9 isolado a quantidade de movimento antes e depois do choque \u00e9 a mesma\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=1250V\u00a0 —\u00a0 V=20m\/s\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo dde V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019 usando o coeficiente de restitui\u00e7\u00e3o\u00a0 — e=\u2502V<\/b><\/span><\/span><\/span>relativa de afastamento<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2502\/\u2502V<\/b><\/span><\/span><\/span>relativa de aproxima\u00e7\u00e3o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2502\u00a0 —\u00a0 0,6=(V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019 \u2013 12)\/(20 \u2013 10)\u00a0 —\u00a0 V\u00ad<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019=18m\/s\u00a0 —\u00a0 em toda colis\u00e3o a quantidade de movimento total sempre se conserva se conserva\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"\u00a0 —\u00a0\u00a0\"\"\"\"\u00a0\u00a0—\u00a0\u00a0\"\"\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

(V) A velocidade do corpo B ap\u00f3s o choque \u00e9 18 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

(V) A massa do corpo A vale 2 kg.<\/b><\/span><\/span><\/span><\/p>\n

(F) O choque \u00e9 perfeitamente el\u00e1stico, pois os dois corpos t\u00eam massas iguais a 2 kg\u00a0 —\u00a0 se o choque \u00e9 el\u00e1stico e = 1.<\/b><\/span><\/span><\/span><\/p>\n

(F) A quantidade de movimento depois do choque \u00e9 menor do que antes do choque\u00a0 —\u00a0 em todo choque a quantidade de movimento total se conserva.<\/b><\/span><\/span><\/span><\/p>\n

(F) A energia dissipada, igual \u00e0 diferen\u00e7a da energia cin\u00e9tica antes do choque e da energia cin\u00e9tica depois do choque, \u00e9 de 64 J\u00a0 —\u00a0 a energia dissipada vale 32J.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (V,V,F,F,F)<\/b><\/span><\/span><\/span><\/p>\n

40-\u00a0Supondo o choque perfeitamente el\u00e1stico e, como eles possuem a mesma massa, ap\u00f3s o choque eles trocam suas velocidades\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=4.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=8J\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0a) Velocidade de A imediatamente antes de se chocar com B\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 mgh=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 10.0,8=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=4m\/s\u00a0 —\u00a0 velocidade de B imediatamente ap\u00f3s o choque com A\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3.4=3.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=4m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) Como, no o atrito \u00e9 desprez\u00edvel o bloco B incide na mola com velocidade de 4m\/s e a mola \u00e9 comprimida de x, at\u00e9 o bloco B parar (V=0)\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 imediatamente antes de se chocar com a mola o bloco s\u00f3 possui energia cin\u00e9tica E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3.16\/2=24J\u00a0 —\u00a0 quando o bloco B p\u00e1ra ele s\u00f3 possui energia potencial el\u00e1stica armazenada\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=4x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=2x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 24=2x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 x=\u221a12\u00a0 —\u00a0 x\u22483,46m<\/b><\/span><\/span><\/span><\/p>\n

c) Agora, com atrito no plano horizontal a \u00fanica for\u00e7a na dire\u00e7\u00e3o do movimento \u00e9 a for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=\u03bcP=0,4.30\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0 12=3.a\u00a0 —\u00a0 a=4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 como a velocidade do bloco B est\u00e1 diminuindo essa acelera\u00e7\u00e3o \u00e9 negativa\u00a0 —\u00a0 equa\u00e7\u00e3o de Torricelli at\u00e9 o bloco B parar\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>\u00a0+ 2.a.\u0394S\u00a0 —\u00a0 0=4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-4).\u0394S\u00a0 —\u00a0 \u0394S=2m\u00a0 —\u00a0\u00a0como d=3m, o bloco B n\u00e3o comprime a mola parando a 1m da mesma.<\/b><\/span><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0Trata-se de uma colis\u00e3o perfeitamente el\u00e1stica com esferas de mesma massa onde elas trocam suas velocidades\u00a0 —\u00a0 o coeficiente de restitui\u00e7\u00e3o vale e=1\u00a0 —\u00a0 a energia cin\u00e9tica se conserva (as velocidades s\u00e3o iguais antes e depois da colis\u00e3o)\u00a0 —\u00a0 em qualquer tipo de colis\u00e3o a quantidade de movimento sempre se conserva\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

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

0. – Quantidade de movimento do sistema antes da colis\u00e3o\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sa<\/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>=2×5 + 8×0=10kg.m\/s\u00a0 —\u00a0 quantidade de movimento do sistema quando se movem juntos com velocidade V e massa m=2 + 8=10kg\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>=mV=10V\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

10=10V\u00a0 —\u00a0 V=1m\/s\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/span><\/p>\n

1. Correta\u00a0 —\u00a0 se movem juntos ap\u00f3s a colis\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

2. Correta\u00a0 —\u00a0 o sistema \u00e9 conservativo (os atritos s\u00e3o desprezados)<\/b><\/span><\/span><\/span><\/p>\n

3. Falsa\u00a0 —\u00a0 quem se conserva \u00e9 a quantidade de movimento do sistema e n\u00e3o de cada bloco.<\/b><\/span><\/span><\/span><\/p>\n

4. Observe a figura abaixo onde os dois blocos juntos tem no ponto P energia mec\u00e2nica E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 e no ponto de altura m\u00e1xima<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Q eles tem energia mec\u00e2nica E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>=mgh\u00a0 —\u00a0 pelo teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10.10.h\u00a0 —\u00a0 h=0,05m=5cm\u00a0 —\u00a0 correta\u00a0<\/b><\/span><\/span><\/span><\/p>\n

I (0,1,2 e 4)\u00a0\u00a0 II (3)<\/b><\/span><\/span><\/span><\/p>\n

44-<\/b><\/span><\/span><\/span> R- B\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

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

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

Observe o esquema abaixo que ilustra a situa\u00e7\u00e3o apresentada\u00a0 —\u00a0 c\u00e1lculo de V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 e g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, considerando<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

desprez\u00edvel o atrito com o ar\u00a0 —\u00a0 aplicando a equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.h\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0 + 2.10.1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20\u00a0 —\u00a0 c\u00e1lculo de V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0na primeira colis\u00e3o com o solo\u00a0 —\u00a0 \u03b5=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 0,8= V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/20\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=16\u00a0 —\u00a0 ela sobe com velocidade V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, atinge a altura m\u00e1xima, retorna e atinge novamente o solo com velocidade V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e, neste segundo choque retorna com velocidade V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u03b5= V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/ V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 0,8= V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/16\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=12,8\u00a0 —\u00a0 ela sobe com velocidade V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>, atinge a altura m\u00e1xima, retorna e atinge novamente o solo com velocidade V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0e, neste terceiro choque retorna com velocidade V<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u03b5= V<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/ V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 0,8= V<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/12,8\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=10,24\u00a0 —\u00a0 c\u00e1lculo, por Torricelli, da altura h<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0atingida ap\u00f3s a terceira colis\u00e3o\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

0= (10,24)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-10).h<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=0,51m\u00a0 —\u00a0\u00a0R- D\u00a0\u00a0—\u00a0 na realidade o coeficiente de restitui\u00e7\u00e3o \u00e9 fornecido por \u03b5=m\u00f3dulo da velocidade relativa antes\/m\u00f3dulo da velocidade relativa depois.<\/b><\/span><\/span><\/span><\/p>\n

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

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

a<\/b><\/span><\/span>) 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 Qantes=M.Vo\u00a0+ ma.Va=9000.80 +1000.0\u00a0 —\u00a0 Qantes=720000kg.m\/s\u00a0 —\u00a0 Qdepois=(M + ma).V=10.000.V<\/b><\/span><\/span><\/p>\n

Qdepois=10000V\u00a0 —\u00a0 Qantes\u00a0= Qdepois\u00a0 —\u00a0 720000=10000V\u00a0 —\u00a0\u00a0V=72km\/h.<\/b><\/span><\/span><\/p>\n

b) Pela figura fornecida voc\u00ea pode determinar a intensidade da for\u00e7a lateral\u00a0\"\"\u00a0\u00a0—\u00a0 sen3o=FL\/Fat\u00a0 —\u00a0 0,05=FL\/8000\u00a0 —\u00a0 FL=400N\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A acelera\u00e7\u00e3o lateral\u00a0\"\"\u00a0do carro tem intensidade\u00a0 —\u00a0 FL=m.aL\u00a0 —\u00a0 400=1600.aL\u00a0 —\u00a0\u00a0aL=0,25m\/s2.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

\u00a0<\/span>47-<\/b><\/span><\/span><\/span><\/p>\n

\u00a0a) Sendo os atritos com a pista desprez\u00edveis o sistema \u00e9 conservativo\u00a0 —\u00a0 chamando de A a esfera da parte superior e de B a da inferior\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=m\u00ad<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=m\u00a0 —\u00a0 inicialmente A encontra-se numa altura h igual ao raio da depress\u00e3o esf\u00e9rica\u00a0 —\u00a0 h=20cm=0,2m\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

utilizando o teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica (figura 1)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mantes<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cantes\u00a0<\/b><\/span><\/span><\/sub><\/span>+ E<\/b><\/span><\/span><\/span>pantes<\/b><\/span><\/span><\/sub><\/span>\u00a0= m.V<\/b><\/span><\/span><\/span>A<\/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.10.0,2\u00a0 —E<\/b><\/span><\/span><\/span>mantes<\/b><\/span><\/span><\/sub><\/span>=2m\u00a0 —\u00a0 depois (V\u2019<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2013 velocidade da esfera A instantes antes de atingir a esfera B)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mdepois<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cdepois<\/b><\/span><\/span><\/sub><\/span>\u00a0+<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>pdepois<\/b><\/span><\/span><\/sub><\/span>=mV\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h= mV\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mdepis<\/b><\/span><\/span><\/sub><\/span>= mV\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mantes<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mdepois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2m = mV\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 = 2m\/s\u00a0 —\u00a0 agora voc\u00ea tem o choque entre as duas esferas\u00a0 —\u00a0 sendo o sistema constitu\u00eddo apenas pelas duas esferas ele \u00e9 isolado e voc\u00ea pode utilizar a conserva\u00e7\u00e3o da quantidade de movimento (figura 2)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.V\u2019<\/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>=m.2 + m.0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=2m\u00a0 —\u00a0 como a colis\u00e3o \u00e9 inel\u00e1stica, as duas esferas obrigatoriamente se movem juntas ap\u00f3s a mesma, com velocidade V\u2019\u2019\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>= 2mV\u2019\u2019\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2=2V\u2019\u00a0 —\u00a0\u00a0V\u2019=1m\/s\u00a0(velocidade das duas esferas se movendo unidas ap\u00f3s a colis\u00e3o)<\/b><\/span><\/span><\/span><\/p>\n

b) C\u00e1lculo da altura m\u00e1xima h que as duas esferas se movendo juntas (massa 2m) atingem e, nesse instante possuem velocidade<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

nula\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mantes<\/b><\/span><\/span><\/sub><\/span>=2mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2m.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mantes<\/b><\/span><\/span><\/sub><\/span>=m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mdepois<\/b><\/span><\/span><\/sub><\/span>=2m.g.h=20mh\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mantes<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>mdepois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m=20mh\u00a0 —\u00a0 h=0,05m\u00a0 —\u00a0 colocando as for\u00e7as que agem sobre as\u00a0 esferas quando elas est\u00e3o em repouso na altura h=0,05m\u00a0 —\u00a0 em todo movimento circular existe uma for\u00e7a resultante denominada for\u00e7a resultante centr\u00edpeta de dire\u00e7\u00e3o radia, sentido para o centro da circunfer\u00eancia e de intensidade F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=M.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 na figura, est\u00e3o colocadas todas as for\u00e7as que agem<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sobre as esferas nessa situa\u00e7\u00e3o\u00a0 —\u00a0 observe que F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=N \u2013 2Pcos\u03b8=2mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=2m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=0\u00a0 —\u00a0 N \u2013 2mg=0\u00a0 —\u00a0 N \u2013 2.0,1.(0,15\/0,2)<\/b><\/span><\/span><\/span><\/p>\n

N = 1,5N<\/b><\/span><\/span><\/span><\/p>\n

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

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

\u00a0\u00a0Observe que, pelo enunciado o sistema \u00e9 mecanicamente conservativo, pois as for\u00e7as de atrito com a rampa e com o ar s\u00e3o desprezadas\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

–\u00a0 c\u00e1lculo da velocidade V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0da part\u00edcula A, abandonada em P, imediatamente antes de colidir com B (ponto Q)\u00a0 —\u00a0 teorema da<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

conserva\u00e7\u00e3o de energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgH=0 + 10mH\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=10mH\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgH=mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10mH=mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20H\u00a0 —\u00a0\u00a0 c\u00e1lculo da<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

velocidade de retorno de A ap\u00f3s o choque perfeitamente el\u00e1stico de coeficiente de restitui\u00e7\u00e3o e=1\u00a0 —\u00a0 e=(m\u00f3dulo da velocidade relativa depois)\/(m\u00f3dulo da velocidade relativa antes)\u00a0 —\u00a0 1=(V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>)\/V<\/b><\/span><\/span><\/span>1\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub><\/span>(I)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>= -mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+2mV<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= – mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2mV<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= – V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0(II)\u00a0 —\u00a0 (I) em (II)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= – V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2(V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= – V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 2V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/3 (a bola A retorna com velocidade V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/3)\u00a0 —\u00a0 por \u00faltimo a esfera A retorna \u00e0 rampa atingindo uma altura m\u00e1xima h no ponto R, quando V=0\u00a0 —\u00a0 teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=m.[(V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>]\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/18\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=0 + mgh\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>=10mh\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/18=10mh\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=180h\u00a0 —\u00a0 20H=180h\u00a0 —\u00a0 h=H\/9\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/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 Colis\u00f5es Mec\u00e2nicas 01- a) e=Vrdepois\/Vrantes\u00a0 —\u00a0 1=(10 + V\u2019Y)\/(5 + 10)\u00a0 —\u00a0 15=10 +V\u2019y\u00a0\u00a0—\u00a0 V\u2019y=5m\/s b) e=Vrdepois\/Vrantes=(4 \u2013 2)\/(5 \u2013 1)\u00a0 —\u00a0\u00a0e=0,5 (choque parcialmente el\u00e1stico) c) 0.8=V\u2019X\/15\u00a0\u00a0—\u00a0 V\u2019X=8m\/s d) 0,6=(V\u2019X\u00a0\u2013 40)\/(40 + 60)\u00a0 —\u00a0 60=V\u2019X\u00a0\u2013 40\u00a0 —\u00a0\u00a0V\u2019X=100 km\/h e) e=0\/4\u00a0 —\u00a0\u00a0e=0 (choque inel\u00e1stico) f) e=\u221a4\/16\u00a0\u00a0—\u00a0 e=0,25 (choque parcialmente el\u00e1stico) 02–\u00a0Considere o choque perfeitamente<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1401,"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-1405","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1405","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=1405"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1405\/revisions"}],"predecessor-version":[{"id":10814,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1405\/revisions\/10814"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1401"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1405"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}