Đáp án:
\(\begin{array}{l}
{v_2} = 66,667m/s\\
\alpha = 60
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
\frac{1}{2}{m_1}v_1^2 + {m_1}gh = \frac{1}{2}{m_1}{v^2}\\
\frac{1}{2}v_1^2 + 10.20 = \frac{1}{2}{.40^2}\\
{v_1} = 20\sqrt 3 m/s\\
{p_2} = \sqrt {{p^2} + p_1^2} \\
{m_2}{v_2} = \sqrt {{{(m{v_0})}^2} + {{({m_1}{v_1})}^2}} \\
(0,8 - 0,5){v_2} = \sqrt {{{(0,8.12,5)}^2} + {{(0,5.20\sqrt 3 )}^2}} \\
{v_2} = 66,667m/s\\
\tan \alpha = \frac{{{p_1}}}{p} = \frac{{0,5.20\sqrt 3 }}{{0,8.12,5}}\\
\alpha = 60
\end{array}\)