Oi, Ismael
A ideia aqui é usar que o sistema é isolado de forças externas.
Segundo a direção [tex3]\text{Oy}, \,[/tex3]
podemos escrever:
[tex3]\begin{array}{} \vec{\text{Q}}_{\text{final}} = \vec{\text{Q}}_{\text{inicial}} \,\,\,\, \Rightarrow \,\,\,\, \vec{\text{Q'}}_{\text{y }} = \vec{\text{Q}}_{\text{y 1}} + \vec{\text{Q}}_{\text{y 2}} \,\,\,\, \Rightarrow \,\,\,\, \(\text{m}_{\text{ 1}} + \text{m}_{\text{2}} \)\cdot \text{v'}_{\text{y }} = 0 + \text{m}_{\text{ 2}} \cdot \text{v}_{\text{2 y }} \,\,\,\, \Rightarrow \,\,\,\, \\\\
\(21600 \)\cdot \text{v'}_{\text{ }} \cdot \cos 30^{\circ} = 19800 \cdot \text{v}_{\text{2}} \,\,\,\, \Rightarrow \,\,\,\, \boxed{ \text{v}_{\text{2}} = \frac{ 21600 \cdot \text{v'} \cdot \cos 30^{\circ} }{19800} }\quad \quad {\color{red} \text{(I)} }
\end{array}[/tex3]
Segundo a direção [tex3]\text{Ox}, \,[/tex3]
podemos escrever:
[tex3]\begin{array}{}\vec{\text{Q}}_{\text{final}} = \vec{\text{Q}}_{\text{inicial}} \,\,\,\, \Rightarrow \,\,\,\, \vec{\text{Q'}}_{\text{x }} = \vec{\text{Q}}_{\text{x 1}} + \vec{\text{Q}}_{\text{x 2}} \,\,\,\, \Rightarrow \,\,\,\, \(\text{m}_{\text{ 1}} + \text{m}_{\text{2}} \)\cdot \text{v'}_{\text{x }} = \text{m}_{\text{ 1}} \cdot \text{v}_{\text{1 x }} + 0 \,\,\,\, \Rightarrow \,\,\,\, \\\\
\(21600 \)\cdot \text{v'}_{\text{ }} \cdot \sen 30^{\circ} = 1800 \cdot 30 \,\,\,\, \Rightarrow \,\,\,\, \text{v'}_{\text{ }} = \frac{ 1800 \, \cdot \, 30 }{21600 \, \cdot \, 0,5 } \,\,\,\, \therefore \,\,\,\, \boxed{\text{v'}_{\text{ }} = 5 \, \text{m/s}} \quad \quad {\color{red} \text{(II)} }
\end{array}[/tex3]
Substituindo o resultado de [tex3]{\color{red} \text{(II)} }[/tex3]
em [tex3]{\color{red} \text{(I)}}, \,[/tex3]
obtemos:
[tex3]\text{v}_{\text{2}} = \frac{ 21600 \cdot \text{v'} \cdot \cos 30^{\circ} }{19800} \,\,\,\, \Rightarrow \,\,\,\, \text{v}_{\text{2}} = \frac{ 21600 \cdot 5 \cdot \cos 30^{\circ} }{19800} \,\,\,\, \therefore \,\,\,\, \boxed{\text{v}_{\text{2}} = 4,72 \, \text{m/s}}[/tex3]