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\begin{document}

\frame{\frametitle{\b Center of Mass (CM) and Momentum - 1} \bigskip

{\bf CM of Earth-Moon (1):} 
$$ x_{\rm cm} = \frac{m_1\,x_1+m_2\,x_2}{m_1+m_2}\,.$$
Choose $x_{\rm cm}=0$ and $m_1$ to be the larger mass.
Then $x=x_2-x_1$ given allows to solve for $x_1$ (negative).
Enter the absolute value of $x_1$.
\smallskip

{\bf Point masses in a plane: CM (2}.
$$x_{\rm cm}=\frac{1}{M}\sum_{i=1}^3m_i\,x_i\,,~~
  y_{\rm cm}=\frac{1}{M}\sum_{i=1}^3m_i\,y_i\,,~~M=\sum_{i=1}^3 m_i\,.$$

{\bf CM of three equal Masses (3):} 
$$ \frac{1}{3}\,(a_1+a_2+a_3)\,\hat{i} +
   \frac{1}{3}\,(b_1+b_2+b_3)\,\hat{j}\,.$$
Then: \vskip -20pt
$$ \vec{v} = \frac{d\,\vec{r}}{d\,t}\,,~~~
   \vec{a} = \frac{d\,\vec{v}}{d\,dt}\,.$$
\vskip 20pt }


\frame{\frametitle{\b Center of Mass (CM) and Momentum - 2} \bigskip

{\bf Nonuniform Rod (4):}  
$$ M_{\rm rod} = \int_0^L dx\,\mu(x) = ?\,,~~~x_{\rm cm} = 
   \frac{1}{M_{\rm rod}} \int_0^L dx\,\mu(x)\,x = ?  $$
\smallskip

{\bf Two Blocks on a Compressed Spring (5):} 
$$ m_1\,v_1=m_2\,v_2\,,~~~\frac{k}{2}\,(\triangle x)^2 =
   m_1\,\frac{v_1^2}{2}+m_2\,\frac{v_2^2}{2}\,.$$
\smallskip

{\bf Impulse on a Golf Ball (6):} 
$$ p = m\,v\,,~~~f = m\,\frac{v}{t}\,. $$
% \bigskip
\smallskip

{\bf Space Walker (7).} Sign convention: $v$ negative.
$$ m\,v = m_1\,v_1+m_2\,v_2\,. $$

\vskip 20pt }

\end{document}
