Question

A satellite that is spinning clockwise has four low-mass solar panels sticking out as shown. A tiny meteor traveling at high speed rips through one of the solar panels and continues in the same direction but at reduced speed. Afterward, calculate the ${\mathbf{v}}_{\mathbf{x}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{y}}$components of the center of mass velocity of the satellite. In Figure 3.64 $\overrightarrow{{\mathbf{v}}_{\mathbf{1}}}\mathbf{}\mathbf{and}\mathbf{}\overrightarrow{{\mathbf{v}}_{\mathbf{2}}}$ are the initial and final velocities of the meteor, and$\overrightarrow{\mathbf{v}}$ is the initial velocity of the center of the mass of the satellite, in the x-direction.

Short answer

the horizontal and vertical velocities of the satellite after the collision are ${\mathrm{v}}_{\mathrm{x}}=\mathrm{v}+\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{cos\theta }$and ${\mathrm{v}}_{\mathrm{y}}=\mathrm{v}+\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{sin\theta }$, respectively.

Step-by-step solution

Identification of given data

The mass of the space junk is m

The mass of the satellite is M

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

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

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

Conservation of linear momentum

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,

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

Conservation of momentum in the x-direction

Horizontal components of the meteor velocities are ${\mathrm{v}}_1\mathrm{cos\theta }\mathrm{and}{\mathrm{v}}_2\mathrm{cos\theta }$, 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,

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

Conservation of momentum in the y-direction

Vertical components of the meteor velocities are ${\mathrm{v}}_1\mathrm{sin\theta }\mathrm{and}{\mathrm{v}}_2\mathrm{sin\theta }$, 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,

$$\begin{gathered} {\mathrm{mv}}_1\mathrm{sin\theta }+\mathrm{M}\times 0={\mathrm{mv}}_2\mathrm{sin\theta }+{\mathrm{Mv}}_{\mathrm{y}} \\ \mathrm{m}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{sin\theta }={\mathrm{Mv}}_{\mathrm{y}} \\ {\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{sin\theta } \end{gathered}$$

Thus, the horizontal and vertical velocities of the satellite after the collision are ${\mathrm{v}}_{\mathrm{x}}=\mathrm{v}+\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{cos\theta }$and ${\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\mathrm{sin\theta }$, respectively.