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\begin{document}

\frame{\frametitle{\b Momentum Conservation - 1} \bigskip

{\bf Mass Mass-Spring Collision (1):} 
$$ m_1\,v_1 = M\,v\,,~~~K_i=m_1\,\frac{v_1^2}{2}\,~~K_f=M\,\frac{v^2}{2}\,,
   ~~~\frac{K}{2}\,(\triangle x)^2 = K_f\, .$$
% \smallskip

{\bf Freight Car-Caboose Collision (2).} Equate the energies and use
momentum conservation:
$$ (1-p)\,m_1\,\frac{v_1^2}{2} = (m_1+m_2)\,\frac{v^2}{2}\,,~~
   v = \frac{m_1\,v_1}{m_1+m_2}\,.$$
Divide common factors out.
\bigskip

{\bf Ball-Bat Collision (3).} As $m_{\rm bat}\gg M=m_{\rm ball}$, the 
rest frame of the bat can be taken as CM frame in which the elastic
collision takes place, and the bat acts like a brick wall:
$$ v'_i=?v_i\,,~~~M\,v'_i = - M\,v'_f\,.$$
Then, transfer $v'_f$ to $v_f$, the final velocity of the ball in the
original frame. Part~2: $v_f^2/v_i^2$.
\vskip 20pt }


\frame{\frametitle{\b Momentum Conservation - 2} \bigskip

{\bf Collison Between Two Balls (4).} Steel balls:
$$ h = r-r\,\cos(\theta)\,. $$
Putty balls ($v_f=?v_i$ from momentum conservation):
$$ K_i=m\,\frac{v_i^2}{2}=M\,h\,g\,,~~K_f=m\,\frac{v_f^2}{2}\,
      =2\,M\,H\,g~~\Rightarrow~~\frac{H}{h}\,.$$
\smallskip

{\bf Glancing Collision (5):} 
\begin{eqnarray} \nonumber
  p_y^i &=& 0 = \sin(\theta_1)\,m_1\,v^f_1+sin(\theta_2)\,m_2\,v^f_2\,,\\
  p_x^i &=& m_1\,v^i = \cos(\theta_1)\,m_1\,v^f_1+\cos(\theta_2)\,m_2\,v^f_2\,,
\end{eqnarray}
with $\theta_2<0$, $m_1=m$ and $m_2=2m$. Eliminate $v_1^f$ in favor of 
$v^f_2$ and solve for $v^f_2$ with $v^i$ given.
\smallskip
\vskip 20pt }

\frame{\frametitle{\b Momentum Conservation - 3} \bigskip

{\bf Elastic Head-On Collision (6):} 
$$ m_1\,v^i_1+m_2\,v^i_2 = m_1\,v^f_1+m_2\,v^f_2\,, $$
$$ \frac{1}{2}\,m_1\,(v^i_1)^2+\frac{1}{2}\,m_2\,(v^i_2)^2 =
   \frac{1}{2}\,m_1\,(v^f_1)^2+\frac{1}{2}\,m_2\,(v^f_2)^2\,. $$
Example (you may have different numbers): $m_1=m$, $v_1^i=4v$, $m_2=2m$ 
and $v_2^i=-v$. Use the first equation to eliminate $(v_2^f)^2$ in the 
second. Solve the resulting quadratic equation for $v_1^f$. Which of 
the solutions to choose? 
Enter the speed which positive by definition. \bigskip

{\bf Two Unequal Mass Balls (7):} Meant is the situation where the
heavy ball bounces back with velocity $-v$, while the lighter
ball is 

still heading towards it with velocity $+v>0$. Thus, the
collission is reduced to the approach of the previous problem.
Once $v_1^f$ is found use
$$ \frac{h'}{h} = \frac{(v^f_1)^2}{v^2}\,.$$

\vskip 20pt }


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