9-57 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_i&=v\\ M&=m_A+m_B=m\\ m_A&=4m_B\\ v_B&=0\\ \end{cases} $$ $$ \begin{cases} m_A&=\frac{4}{5}M\\ m_B&=\frac{1}{5}M\\ \end{cases} $$ $$\Delta \Sigma \vec p=0,$$ $$ \begin{aligned} Mv_i&=m_Av_A+m_Bv_B\\ &=m_Av_A \end{aligned} $$ $$ \begin{aligned} v_A&=\frac{M}{m_A}v_i\\ &=\frac{5m_A}{m_A}v_i\\ &=5v_i\\ \end{aligned} $$ $$ \begin{aligned} \Delta \Sigma \KE&=\Delta \sum \(\frac{1}{2}mv^2\)\\ &=\(\frac{1}{2}m_A{v_A}^2+\frac{1}{2}m_B{v_B}^2\)-\frac{1}{2}M{v_i}^2\\ &=\frac{1}{2}\(\frac{4}{5}M\)(5v_i)^2-\frac{1}{2} M{v_i}^2\\ &=\frac{1}{2}M{v_i}^2\(\frac{4}{5}\cdot 5^2-1\)\\ &=19\cdot\frac{1}{2}M{v_i}^2 \end{aligned} $$ $$\Ans=19\KE_i=19\cdot\frac{1}{2}mv^2$$