Oi, Ismael
Como o choque entre as partículas constitui um sistema isolado de forças externas, podemos aplicar o
Princípio da Conservação da Quantidade de Movimento. Segundo a direção [tex3]\text{Oy}, \,[/tex3]
podemos escrever:
[tex3]\vec{\text{Q}}_{\text{final}} = \vec{\text{Q}}_{\text{inicial}} \,\,\,\, \Rightarrow \,\,\,\, \vec{\text{Q'}}_{\text{y a}} + \vec{\text{Q'}}_{\text{y b}} = \vec{\text{Q}}_{\text{y a}} + \vec{\text{Q}}_{\text{y b}} \,\,\,\, \Rightarrow \,\,\,\, \text{m}_{\text{ a}} \cdot \text{v'}_{\text{y a}} + \text{m}_{\text{b}} \cdot \text{v'}_{\text{y b}} = 0 + 0 \\\\
\text{m}_{\text{ a}} \cdot \text{v'}_{\text{a}} \cdot \sen(\alpha) + \text{m}_{\text{b}} \cdot \text{v'}_{\text{b}} \cdot \sen (\beta) = 0 \,\,\,\, \Rightarrow \,\,\,\, 2 \cdot 10 \cdot 0,6 + 12 \cdot \text{v}_{\text{b}} \cdot \sen (\beta) = 0 \quad \quad {\color{red} \text{(I)} }
[/tex3]
Segundo a direção [tex3]\text{Ox}, \,[/tex3]
podemos escrever:
[tex3]\vec{\text{Q}}_{\text{final}} = \vec{\text{Q}}_{\text{inicial}} \,\,\,\, \Rightarrow \,\,\,\, \vec{\text{Q'}}_{\text{x a}} + \vec{\text{Q'}}_{\text{x b}} = \vec{\text{Q}}_{\text{x a}} + \vec{\text{Q}}_{\text{x b}} \,\,\,\, \Rightarrow \,\,\,\, \text{m}_{\text{ a}} \cdot \text{v'}_{\text{x a}} + \text{m}_{\text{b}} \cdot \text{v'}_{\text{x b}} = \text{m}_{\text{a}} \cdot \text{v}_{\text{x a}} + 0 \\\\
\text{m}_{\text{ a}} \cdot \text{v'}_{\text{a}} \cdot \cos (\alpha) + \text{m}_{\text{b}} \cdot \text{v'}_{\text{b}} \cdot \cos (\beta) =\text{m}_{\text{a}} \cdot \text{v}_{\text{x a}} \,\,\,\, \Rightarrow \,\,\,\, 2 \cdot 12 \cdot 0,8 + 12 \cdot \text{v}_{\text{b}} \cdot \cos (\beta) = 2 \cdot 26 \quad \quad {\color{red} \text{(II)} }
[/tex3]
Daí,
[tex3]\begin{cases}
2 \cdot 10 \cdot 0,6 + 12 \cdot \text{v}_{\text{b}} \cdot \sen (\beta) = 0 \quad \quad \quad \,\,\, {\color{red} \text{(I)} } \\
2 \cdot 12 \cdot 0,8 + 12 \cdot \text{v}_{\text{b}} \cdot \cos (\beta) = 2 \cdot 26 \quad \quad {\color{red} \text{(II)} }
\end{cases}[/tex3]
Tente terminar.