Física I ⇒ (IME) Quantidade De Movimento Tópico resolvido

[ismaelmat]

18.262 - (IME) Dois corpos, A e B, de massas diferentes MA e MB, têm a mesma energia cinética. Qual a relação entre os módulos de suas quantidades de movimento?

Resposta

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Gabarito: QA/QB = [tex3]\sqrt{\frac{MA}{MB}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt{\frac{MA}{MB}}[/tex3]

[MateusQqMD]

E aí, Ismael. Vou desenvolver os cálculos e qualquer dúvida você manda aí.

[tex3]1) \,\, \text{E}_{\text{C A}} = \text{E}_{\text{C B}} \,\,\,\, \Rightarrow \,\,\,\, \frac{ \text{m}_{\text{A}} \cdot \text{v}^2_{\text{A}} }{2} = \frac{ \text{m}_{\text{B}} \cdot \text{v}^2_{\text{B}} }{2} \,\,\,\, \Rightarrow \,\,\,\, \frac{ \text{v}^2_{\text{A}} }{\text{v}^2_{\text{B}} } = \frac{\text{m}_{\text{B}} }{\text{m}_{\text{A}} } \,\,\, \therefore \,\,\, \boxed{ \left| \frac{ \text{v}_{\text{A}} }{\text{v}_{\text{B}} } \right|= \sqrt{ \frac{\text{m}_{\text{B}} }{\text{m}_{\text{A}} }}} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]1) \,\, \text{E}_{\text{C A}} = \text{E}_{\text{C B}} \,\,\,\, \Rightarrow \,\,\,\, \frac{ \text{m}_{\text{A}} \cdot \text{v}^2_{\text{A}} }{2} = \frac{ \text{m}_{\text{B}} \cdot \text{v}^2_{\text{B}} }{2} \,\,\,\, \Rightarrow \,\,\,\, \frac{ \text{v}^2_{\text{A}} }{\text{v}^2_{\text{B}} } = \frac{\text{m}_{\text{B}} }{\text{m}_{\text{A}} } \,\,\, \therefore \,\,\, \boxed{ \left| \frac{ \text{v}_{\text{A}} }{\text{v}_{\text{B}} } \right|= \sqrt{ \frac{\text{m}_{\text{B}} }{\text{m}_{\text{A}} }}} [/tex3]

[tex3]2) \,\,[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]2) \,\,[/tex3]

Sabemos que [tex3]\text{Q}_{\text{A}} = \text{m}_{\text{A}} \cdot \text{v}_{\text{A}} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\text{Q}_{\text{A}} = \text{m}_{\text{A}} \cdot \text{v}_{\text{A}} [/tex3]

e [tex3]\text{Q}_{\text{B}} = \text{m}_{\text{B}} \cdot \text{v}_{\text{B}},[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\text{Q}_{\text{B}} = \text{m}_{\text{B}} \cdot \text{v}_{\text{B}},[/tex3]

logo:

[tex3]\left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \frac{ \text{m}_{\text{A}} \cdot \text{v}_{\text{A}} }{ \text{m}_{\text{B}} \cdot \text{v}_{\text{B}} } \,\,\,\, \Rightarrow \,\,\,\, \left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \frac{ \text{m}_{\text{A}} }{ \text{m}_{\text{B}} } \cdot \sqrt{ \frac{\text{m}_{\text{B}} }{\text{m}_{\text{A}} }} \,\,\,\, \Rightarrow \,\,\,\, \left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \sqrt{ \frac{\text{m}_{\text{A}}^2\cdot \text{m}_{\text{B}} }{\text{m}_{\text{B}}^2 \cdot \text{m}_{\text{A}} }} \,\,\, \therefore \,\,\, \boxed{ \left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \sqrt{ \frac{\text{m}_{\text{A}} }{\text{m}_{\text{B}} }}} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \frac{ \text{m}_{\text{A}} \cdot \text{v}_{\text{A}} }{ \text{m}_{\text{B}} \cdot \text{v}_{\text{B}} } \,\,\,\, \Rightarrow \,\,\,\, \left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \frac{ \text{m}_{\text{A}} }{ \text{m}_{\text{B}} } \cdot \sqrt{ \frac{\text{m}_{\text{B}} }{\text{m}_{\text{A}} }} \,\,\,\, \Rightarrow \,\,\,\, \left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \sqrt{ \frac{\text{m}_{\text{A}}^2\cdot \text{m}_{\text{B}} }{\text{m}_{\text{B}}^2 \cdot \text{m}_{\text{A}} }} \,\,\, \therefore \,\,\, \boxed{ \left| \frac{ \text{Q}_{\text{A}} }{ \text{Q}_{\text{B}} } \right| = \sqrt{ \frac{\text{m}_{\text{A}} }{\text{m}_{\text{B}} }}} [/tex3]

[MateusQqMD]

Eu não representei os vetores porque os códigos já estavam grandes demais..