Đáp án:

 \(\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{8}\)

Giải thích các bước giải:

Áp dụng bảo toàn động lượng:

\(\begin{array}{l}
{m_1}{v_1} + {m_2}{v_2} = {m_1}{v_1}' + {m_2}{v_2}'\\
 \Rightarrow {m_1}.4 = {m_1}.2{v_2}' - {m_2}{v_2}'\\
 \Rightarrow 4{m_1} = {v_2}'\left( {2{m_1} - {m_2}} \right)\\
 \Rightarrow {v_2}' = \dfrac{{4{m_1}}}{{2{m_1} - {m_2}}}\\
 \Rightarrow {v_1}' = \dfrac{{8{m_1}}}{{2{m_1} - {m_2}}}
\end{array}\)

Áp dụng bảo toàn động năng:

\(\begin{array}{l}
\dfrac{1}{2}{m_1}v_1^2 = \dfrac{1}{2}{m_1}{\left( {{v_1}'} \right)^2} + \dfrac{1}{2}{m_2}{\left( {{v_2}'} \right)^2}\\
 \Rightarrow {m_1}{.4^2} = {m_1}.{\left( {\dfrac{{8{m_1}}}{{2{m_1} - {m_2}}}} \right)^2} + {m_2}{\left( {\dfrac{{8{m_1}}}{{2{m_1} - {m_2}}}} \right)^2}\\
 \Rightarrow 16 = \dfrac{{64m_1^2}}{{{{\left( {2{m_1} - {m_2}} \right)}^2}}} + \dfrac{{64{m_1}{m_2}}}{{{{\left( {2{m_1} - {m_2}} \right)}^2}}}\\
 \Rightarrow 16\left( {4m_1^2 - 4{m_1}{m_2} + m_2^2} \right) = 64m_1^2 + 64{m_1}{m_2}\\
 \Rightarrow 128{m_1}{m_2} - 16m_2^2 = 0\\
 \Rightarrow {m_2} = 8{m_1} \Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{8}
\end{array}\)