9-38 할리데이 11판 솔루션 일반물리학

$$ \put \begin{cases} 0:\text{Start}\\ 1:\text{After Turn}\\ 2:\text{Stop} \end{cases} $$ $$ \begin{cases} m&=1400\ut{kg}\\ v_0&=5.3\ut{m/s}\\ \vec v_0&=v_0\j\\ \Delta t_{0\rarr1} &= 4.6\ut{s}\\ \vec v_1&=v_0\i\\ \Delta t_{1\rarr2} &= 350\ut{ms}=0.35\ut{s}\\ \vec v_2&=0 \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec J_{0\rarr1} &=\Delta \vec p_{0\rarr1}\\ &=\vec p_1-\vec p_0\\ &=m v_0\i-m v_0\j\\ &=7420\i-7420\j\ut{N\cdot s}\\ &\approx 7.4\i-7.4\j\ut{kN\cdot s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec J_{1\rarr2} &=\Delta \vec p_{1\rarr2}\\ &=\vec p_2-\vec p_1\\ &=0-m v_1\j\\ &=-7420\i\ut{N\cdot s}\\ &\approx -7.4\i\ut{kN\cdot s}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \vec F_{0\rarr1}&=\frac{\vec J_{0\rarr1}}{\Delta t_{0\rarr1}}\\ &=\frac{37100}{23}\i-\frac{37100}{23}\j\ut{N}\\ &\approx 1613.0434782608695\i\\ &~~~~~-1613.0434782608695\j\ut{N}\\ &\approx 1.6\i-1.6\j\ut{kN} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \vec F_{1\rarr2}&=\frac{\vec J_{1\rarr2}}{\Delta t_{1\rarr2}}\\ &=-21200\i\ut{N}\\ &\approx -21\i\ut{kN} \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \vec F_{0\rarr1}&=\frac{37100}{23}\i-\frac{37100}{23}\j\ut{N},\\ \end{aligned} $$ $$\theta_F=-\frac{\pi}{4}\ut{rad}$$