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61P

Expert-verifiedFound in: Page 127

Book edition
4th edition

Author(s)
Ruth W. Chabay, Bruce A. Sherwood

Pages
1135 pages

ISBN
9781118875865

**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.**

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.

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 $\overrightarrow{\mathrm{v}}$

**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 $\mathrm{vcos\theta}$, 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,

$\mathrm{mvcos\theta}={\mathrm{Mv}}_{\mathrm{x}}\phantom{\rule{0ex}{0ex}}{\mathrm{Mv}}_{\mathrm{x}}=\mathrm{mv}\mathrm{cos\theta}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vcos\theta}\right)\phantom{\rule{0ex}{0ex}}$

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,

**$\mathrm{mvsin\theta}={\mathrm{Mv}}_{\mathrm{y}}\phantom{\rule{0ex}{0ex}}{\mathrm{Mv}}_{\mathrm{y}}=\mathrm{mv}\mathrm{sin\theta}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{vsin\theta}\right)$ **

** **

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.

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