Question
1) ( valor 04)(PUC-RJ) Um garoto de massa 30 kg está parado sobre uma grande plataforma de massa 120,kg também em repouso em uma superficle de gelo. Ele começa a correr de 2,0m/s Sabendo que não há atrito entre a plataforma e a superficie de gelo a velocidade horizontalmente para a direita e um observador, fora da plataforma, mede que sua 6 com quê a plataforma se desloca para a esquerda, para esse observador, é, em m/s
Solution
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(189 Votos)
Elisa
Elite · Tutor por 8 anos
Resposta
1. v = 60 kg.m/s ÷ 120 kg = 0.5 m/s.
Explicação
Using the principle of conservation of momentum before and after the boy started running:Initially, both the boy and the platform are at rest, thus the total momentum in the y direction before running is 0 (Because the sum of momentum of the boy and the platform is 0 as both were at rest).After the boy has started to run with a velocity of 2.0m/s to the right, there is now momentum towards the right. However, since the total momentum in the closed system (boy + skateboard) must remain 0 (the same as before running), it means the soulbeard must now has momentum in the left direction to counter and equilibrium the momentum caused by the moving boy.The boy gains a momentum towards the right which can be calculated as Pboy= mass_of_boy x speed_of_boy = 30 kg x 2.0 m/s = 60 kg.m/s.In general, to keep total momentum zero, platform's momentum in opposite direction must also be 60 kg.m/s and as per definition of momentum (P=mv), let's denote the platform's Notre speed as v, we have: Pplatform = mass_of_Platform x speed_of_Platform = 120 kg x v = 60 kg.m/s. We can now solve for v, the speed of the platform.