9-39 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_{1i}&=10\ut{m/s} \end{cases} $$ $$ \begin{cases} \theta_1&=0\\ \theta_2&=+30\degree\\ \theta_3&=-30\degree\\ \end{cases} $$ $$\because \triangle 123=\text{Equilateral Triangle}$$ $$\Delta \Sigma \vec p=0,$$ $$ \begin{aligned} \vec p_{1i}&=\vec p_{1f}+\vec p_{2f}+\vec p_{3f}\\ m\vec v_{1i}&=m\vec v_{1f}+m\vec v_{2f}+m\vec v_{3f}\\ \vec v_{1i}&=\vec v_{1f}+\vec v_{2f}+\vec v_{3f}\\ \end{aligned} $$ $$\Delta \Sigma \KE=0$$ $$ \begin{aligned} \KE_{1i}&=\KE_{1f}+\KE_{2f}+\KE_{3f}\\ \frac{1}{2}m{v_{1i}}^2&=\frac{1}{2}m{v_{1f}}^2+\frac{1}{2}m{v_{2f}}^2+\frac{1}{2}m{v_{3f}}^2\\ {v_{1i}}^2&={v_{1f}}^2+{v_{2f}}^2+{v_{3f}}^2\\ \end{aligned} $$ $$ \begin{cases} v_{1i}&=v_{1fx}+v_{2fx}+v_{3fx}\\ 0&=v_{1fy}+v_{2fy}+v_{3fy}\\ {v_{1i}}^2&={v_{1f}}^2+{v_{2f}}^2+{v_{3f}}^2 \end{cases} $$ $$ \begin{cases} v_{1i}&=v_{1f}\cos\theta_1+v_{2f}\cos\theta_2+v_{3f}\cos\theta_3\\ 0&=v_{1f}\sin\theta_1+v_{2f}\sin\theta_2+v_{3f}\sin\theta_3\\ {v_{1i}}^2&={v_{1f}}^2+{v_{2f}}^2+{v_{3f}}^2 \end{cases} $$ $$ \begin{cases} v_{1f}&=-\cfrac{1}{5}v_{1i}\\ v_{2f}&=\cfrac{2\sqrt3}{5}v_{1i}\\ v_{3f}&=\cfrac{2\sqrt3}{5}v_{1i}\\ \end{cases} $$ $$\ab{a,b,c,d,e,f}$$ $$ \begin{cases} v_{1f}&=-2\ut{m/s},&\theta_1=0\\ v_{2f}&=4\sqrt3\ut{m/s},&\theta_2=+30\degree\\ v_{3f}&=4\sqrt3\ut{m/s},&\theta_3=-30\degree\\ \end{cases} $$ $$ \begin{cases} v_{1f}&=2.0\ut{m/s},&\theta_1=180\degree\\ v_{2f}&\approx 6.928203230275509\ut{m/s},&\theta_2=+30\degree\\ v_{3f}&\approx6.928203230275509\ut{m/s},&\theta_3=-30\degree\\ \end{cases} $$ $$ \begin{cases} v_{1f}&=2.0\ut{m/s},&\theta_1=180\degree\\ v_{2f}&\approx 6.9\ut{m/s},&\theta_2=+30\degree\\ v_{3f}&\approx6.9\ut{m/s},&\theta_3=-30\degree\\ \end{cases} $$