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Found in: Page 127

### Matter & Interactions

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.

See the step by step solution

## Step 1: 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 $\stackrel{\to }{\mathrm{v}}$

## Step 2: 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,

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

## Step 3: 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,

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

## Step 4: 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,

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

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.