Đáp án:

\(\begin{array}{l}
\\
18:\\
a.\\
{v_2} = 848,53m/s\\
\alpha  = 45^\circ \\
b.\\
h = 1125m\\
19.\\
{v_2} = 346,3m/s\\
\alpha  = 30^\circ 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
18:\\
m = 20kg,v = 150m/s\\
{m_1} = 15kg,{v_1} = 200m/s\\
 -  -  -  -  - \\
a.{v_2} = ?\\
\alpha  = ?\\
b.\\
h = ?\\
\\
a.\\
m\vec v = {m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}\\
(\vec v,{{\vec v}_1}) = 90\\
{m_2}{v_2} = \sqrt {{{({m_1}{v_1})}^2} + {{(mv)}^2}} \\
{v_2} = \frac{{\sqrt {{{({m_1}{v_1})}^2} + {{(mv)}^2}} }}{m} = \frac{{\sqrt {{{(15.200)}^2} + {{(20.150)}^2}} }}{{20 - 15}} = 848,53m/s\\
\tan \alpha  = \frac{{mv}}{{{m_1}{v_1}}} = \frac{{20.150}}{{15.200}} = 1\\
\alpha  = 45^\circ \\
b.\\
\frac{1}{2}m{v^2} = mgh\\
\frac{1}{2}{.20.150^2} = 20.10.h\\
h = 1125m\\
19.\\
v = 200m/s\\
{m_1} = 5kg,{v_1} = 346m/s\\
{m_2} = 10kg\\
 -  -  -  -  -  -  -  -  - \\
{v_2} = ?\alpha  = ?\\
m\vec v = {m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}\\
(\vec v,{{\vec v}_1}) = 90\\
{m_2}{v_2} = \sqrt {{{({m_1}{v_1})}^2} + {{(mv)}^2}} \\
{v_2} = \frac{{\sqrt {{{({m_1}{v_1})}^2} + {{(mv)}^2}} }}{m} = \frac{{\sqrt {{{(5.346)}^2} + {{((10 + 5).200)}^2}} }}{{10}} = 346,3m/s\\
\tan \alpha  = \frac{{{m_1}{v_1}}}{{mv}} = \frac{{5.346}}{{15.200}}\\
\alpha  = 30^\circ 
\end{array}\)