9-14 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_1&=16.7\times10^{-27}\ut{kg}\\ m_2&=8.35\times10^{-27}\ut{kg}\\ m_3&=11.7\times10^{-27}\ut{kg}\\ \vec v_1&=+6.00\times10^6\i\ut{m/s}\\ \vec v_2&=-8.00\times10^6\j\ut{m/s}\\ \end{cases} $$ $$\ab{a}$$ $$\Delta \Sigma \vec P=0,$$ $$ \vec P_i=\Sigma \vec P_f $$ $$ 0=m_1\vec v_1+m_2\vec v_2+m_3\vec v_3 $$ $$ \begin{aligned} \vec v_3&=\frac{-m_1\vec v_1-m_2\vec v_2}{m_3}\\ &=-\frac{m_1}{m_3}\vec v_1-\frac{m_2}{m_3}\vec v_2\\ &=\bra{-\frac{16.7}{11.7}\cdot6\i-\frac{8.35}{11.7}\cdot(-8)\j} \times10^6\ut{m/s}\\ &=\(-\frac{334}{39}\i+\frac{668}{117}\j\) \times10^6\ut{m/s}\\ &\approx (-8.564102564102564\i\\ &~~~~~~+5.7094017094017095\j)\times10^6\ut{m/s}\\ &\approx (-8.56\i+5.71\j)\times10^6\ut{m/s}\\ &\approx -8.56\i+5.71\j\ut{Mm/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Delta \KE&=\KE_f-\KE_i\\ &=\KE_f\\ &=\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2+\frac{1}{2}m_3{v_3}^2\\ &=\frac{90013}{14625}\times10^{-13}\ut{J}\\ &\approx 6.154735042735043\times10^{-13}\ut{J}\\ &\approx 6.15\times10^{-13}\ut{J}\\ &\approx 615\times10^{-15}\ut{J}\\ &\approx 615\ut{fJ}\\ \end{aligned} $$