91Ó°ÊÓ

The mass of the space junk is m

The mass of the satellite is M

The initial velocity of the meteor is $\stackrel{→}{{\mathrm{v}}_1}$

The final velocity of the meteor is $\stackrel{→}{{\mathrm{v}}_2}$

The final velocity of the center of mass of the satellite is $\stackrel{→}{\mathrm{v}}$

The linear momentum remains conserved in an elastic collision, therefore, if two masses ${\mathbf{m}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{m}}_{\mathbf{2}}$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,

${\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{2}}\mathbf{=}{\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{3}}\mathbf{+}{\mathbf{m}}_2{\mathbf{v}}_{\mathbf{4}}$

Horizontal components of the meteor velocities are ${\mathrm{v}}_1\mathrm{³¦´Ç²õθ}\mathrm{and}{\mathrm{v}}_2\mathrm{³¦´Ç²õθ}$, respectively, while the center of mass of the satellite velocities are $\mathrm{v}\mathrm{and}{\mathrm{v}}_{\mathrm{x}}$, respectively. Therefore, according to the law of conservation of linear momentum, we will have,

$$\begin{gathered} {\mathrm{mv}}_1\mathrm{³¦´Ç²õθ}+\mathrm{Mv}={\mathrm{mv}}_2\mathrm{³¦´Ç²õθ}+{\mathrm{Mv}}_{\mathrm{x}} \\ \mathrm{m}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{³¦´Ç²õθ}+\mathrm{Mv}={\mathrm{Mv}}_{\mathrm{x}} \\ {\mathrm{v}}_{\mathrm{x}}=\mathrm{v}+\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{³¦´Ç²õθ} \end{gathered}$$

Vertical components of the meteor velocities are ${\mathrm{v}}_1\mathrm{²õ¾\pm ²Ôθ}\mathrm{and}{\mathrm{v}}_2\mathrm{²õ¾\pm ²Ôθ}$, respectively, while the center of mass of the satellite velocities are $0\mathrm{and}{\mathrm{v}}_{\mathrm{y}}$, respectively. Therefore, according to the law of conservation of linear momentum, we will have,

$$\begin{gathered} {\mathrm{mv}}_1\mathrm{²õ¾\pm ²Ôθ}+\mathrm{M}×0={\mathrm{mv}}_2\mathrm{²õ¾\pm ²Ôθ}+{\mathrm{Mv}}_{\mathrm{y}} \\ \mathrm{m}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{²õ¾\pm ²Ôθ}={\mathrm{Mv}}_{\mathrm{y}} \\ {\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{²õ¾\pm ²Ôθ} \end{gathered}$$

Thus, the horizontal and vertical velocities of the satellite after the collision are ${\mathrm{v}}_{\mathrm{x}}=\mathrm{v}+\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{³¦´Ç²õθ}$and ${\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{²õ¾\pm ²Ôθ}$, respectively.