Question

A space station has the form of a hoop of radius R, with mass M. Initially its center of mass is not moving, but it is spinning. Then a small package of mass m is thrown by a spring-loaded gun toward a nearby spacecraft as shown in Figure 3.66; the package has a speed v after launch. Calculate the center-of-mass velocity (a vector) of the space station after the launch.

Short answer

The horizontal and vertical components of the center of the mass of the satellite are ${\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vcos\theta }\right)$and ${\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vsin\theta }\right)$, respectively.

Step-by-step solution

Given

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 $\overrightarrow{\mathrm{v}}$

Conservation of linear momentum

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

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

Conservation of momentum in the x-direction

Horizontal components of the launched package are $\mathrm{vcos\theta }$, 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,

$$\begin{gathered} \mathrm{mvcos\theta }={\mathrm{Mv}}_{\mathrm{x}} \\ {\mathrm{Mv}}_{\mathrm{x}}=\mathrm{mv}\mathrm{cos\theta } \\ {\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vcos\theta }\right) \end{gathered}$$

Conservation of momentum in the y-direction

Vertical components of the launched package are $\mathrm{vsin\theta }$, 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,

$$\begin{gathered} \mathrm{mvsin\theta }={\mathrm{Mv}}_{\mathrm{y}} \\ {\mathrm{Mv}}_{\mathrm{y}}=\mathrm{mv}\mathrm{sin\theta } \\ {\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vsin\theta }\right) \end{gathered}$$

Thus, the horizontal and vertical components of the center of the mass of the satellite are ${\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vcos\theta }\right)$and ${\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vsin\theta }\right)$, respectively.