9-6 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=2140\ut{kg}\\ m_B&=242\ut{kg}\\ v_{Ai}&=0\\ v_{Bi}&=5.3\ut{m/s}\\ \end{cases} $$ $$\Delta \Sigma \vec P=0,$$ $$\vec P_i=\vec P_{Af}+\vec P_{Bf},$$ $$ \begin{cases} m_Bv_i&=m_Av_{Af}+m_Bv_{Bf}\\ v_{B\larr Af}&=v_{Bf}-v_{Af} \end{cases} $$ $$v_{Af}=\frac{m_B}{m_A+m_B}(v_i-v_{B\larr Af})$$ $$\ab{a}$$ $$v_{B\larr Af}=0$$ $$ \begin{aligned} v_{Af}&=\frac{m_B}{m_A+m_B}v_i\\ &=\frac{6413}{11910}\ut{m/s}\\ &\approx 0.5384550797649035\ut{m/s}\\ &\approx 0.54\ut{m/s}\\ &\approx 54\ut{cm/s}\\ \end{aligned} $$ $$\ab{b}$$ $$v_{B\larr Af}=5.3\ut{m/s}=v_i$$ $$ \begin{aligned} v_{Af}&=0 \end{aligned} $$ $$\ab{c}$$ $$v_{B\larr Af}=-5.3\ut{m/s}=-v_i$$ $$ \begin{aligned} v_{Af}&=\frac{2m_B}{m_A+m_B}v_i\\ &=\frac{6413}{5955}\ut{m/s}\\ &\approx 1.076910159529807\ut{m/s}\\ &\approx 1.1\ut{m/s}\\ \end{aligned} $$