Đáp án:
\(\begin{array}{l}
109.\\
{v_1}' = - 1m/s\\
110.\\
{v_1} = 3m/s\\
111.\\
a.\\
v = 3,4m/s\\
b.\\
v = 0,2m/s
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
109.\\
p = p'\\
{m_1}{v_1} = {m_1}{v_1}' + {m_2}{v_2}\\
{v_1}' = \frac{{{m_1}{v_1} - {m_2}{v_2}}}{{{m_1}}} = \frac{{3000.4 - 5000.3}}{3} = - 1m/s
\end{array}\)
v'<0 nên chueeyr động theo hướng ngược lại
\(\begin{array}{l}
110.\\
p = p'\\
{m_1}{v_1} = ({m_1} + {m_2})v\\
{v_1} = \frac{{({m_1} + {m_2})v}}{{{m_1}}} = \frac{{(4 + 2)2}}{4} = 3m/s\\
111.\\
a.\\
p = p'\\
{m_1}{v_1} + {m_2}{v_2} = ({m_1} + {m_2})v\\
v = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}} = \frac{{60.4 + 90.3}}{{60 + 90}} = 3,4m/s\\
b.\\
{m_2}{v_2} - {m_1}{v_1} = ({m_1} + {m_2})v\\
v = \frac{{{m_2}{v_2} - {m_1}{v_1}}}{{{m_1} + {m_2}}} = \frac{{90.3 - 60.4}}{{60 + 90}} = 0,2m/s
\end{array}\)