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The mass of the launched package is m

The mass of the space station is M

The radius of the space station is M

The initial speed of the space station is 0

The initial speed of the launched package is $\stackrel{→}{\mathrm{v}}$

The linear momentum remains conserved in an elastic collision, therefore, if two masses ${\mathbf{m}}_{\mathbf{1}}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{m}}_{\mathbf{2}}$have the initial velocities of ${\mathbf{v}}_{\mathbf{1}}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{2}}$and final velocities of ${\mathbf{v}}_{\mathbf{3}}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{4}}$, then according to the law of conservation of momentum, we will have,

${\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{2}}\mathbf{=}{\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{3}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{4}}$

Horizontal components of the launched package are $\mathrm{\pm ¹³¦´Ç²õθ}$, while the space station has the speeds before and after the launch $0\mathrm{and}{\mathrm{v}}_{\mathrm{x}}$, respectively.

Therefore, according to the law of conservation of linear momentum, we will have,

$$\begin{gathered} \mathrm{m\pm ¹³¦´Ç²õθ}={\mathrm{Mv}}_{\mathrm{x}} \\ {\mathrm{Mv}}_{\mathrm{x}}=\mathrm{mv}\mathrm{³¦´Ç²õθ} \\ {\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{\pm ¹³¦´Ç²õθ}\right) \end{gathered}$$

Vertical components of the launched package are $\mathrm{\pm ¹²õ¾\pm ²Ôθ}$, while the space station has the speeds before and after the launch $0\mathrm{and}{\mathrm{v}}_{\mathrm{y}}$, respectively.

Therefore, according to the law of conservation of linear momentum, we will have,

$$\begin{gathered} \mathrm{m\pm ¹²õ¾\pm ²Ôθ}={\mathrm{Mv}}_{\mathrm{y}} \\ {\mathrm{Mv}}_{\mathrm{y}}=\mathrm{mv}\mathrm{²õ¾\pm ²Ôθ} \\ {\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{\pm ¹²õ¾\pm ²Ôθ}\right) \end{gathered}$$

Thus, the horizontal and vertical components of the center of the mass of the satellite are ${\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{\pm ¹³¦´Ç²õθ}\right)$and ${\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{\pm ¹²õ¾\pm ²Ôθ}\right)$, respectively.