솔루션피아

$$ \put \begin{cases} \KE : \text{Kinetic Energy}\\\end{cases} $$$$ \begin{cases} m&=150\ut{g}\\v_i&=5.2\ut{m/s}\\\KE_f&=\frac{1}{2}\KE_i\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\KE_f&=\frac{1}{2}\KE_i\\\frac{1}{2}m{v_f}^2&=\frac{1}{2}\cdot\frac{1}{2}m{v_i}^2\\\end{aligned} $$$$ \begin{aligned}v_f&=\sqrt\frac{1}{2}\cdot{v_i}\\&=\frac{v_i}{\sqrt2}\\&=\frac{13\sqrt2}{5}\ut{m/s}\..

$$\put \begin{cases} r:\text{red}\\b:\text{blue}\end{cases} $$$$ \begin{cases} v_i&=4.00\ut{m/s}\\\theta&=40.0\degree\\\end{cases} $$$$ \begin{cases} v_{ri}&=v_i\cos\theta\i+v_i\sin\theta\j\\v_{bi}&=v_i\cos\theta\i-v_i\sin\theta\j\\\end{cases} $$$$ \Delta p_x = 0, $$$$ \begin{cases} p_{rix}=p_{rfx}\\p_{bix}=p_{bfx}\\\end{cases} $$$$ \put \begin{cases} v_{\text{in}}&=\vec v_{..

$$ \begin{cases} v_{1i}&=10\ut{m/s} \end{cases} $$ $$ \begin{cases} \theta_1&=0\\ \theta_2&=+30\degree\\ \theta_3&=-30\degree\\ \end{cases} $$ $$\because \triangle 123=\text{Equilateral Triangle}$$ $$\Delta \Sigma \vec p=0,$$ $$ \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..

$$ \put \begin{cases} 0:\text{Start}\\ 1:\text{After Turn}\\ 2:\text{Stop} \end{cases} $$ $$ \begin{cases} m&=1400\ut{kg}\\ v_0&=5.3\ut{m/s}\\ \vec v_0&=v_0\j\\ \Delta t_{0\rarr1} &= 4.6\ut{s}\\ \vec v_1&=v_0\i\\ \Delta t_{1\rarr2} &= 350\ut{ms}=0.35\ut{s}\\ \vec v_2&=0 \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec J_{0\rarr1} &=\Delta \vec p_{0\rarr1}\\ &=\vec p_1-\vec p_0\\ &=m v_0\i-m v_0..

$$ \begin{cases} m_L&=1.00\ut{kg}\\ m_R&=0.500\ut{kg}\\ \end{cases} $$ $$\Delta \Sigma \vec p=0,$$ $$m_Lv_L+m_Rv_R=0$$ $$\ab{a}$$ $$v_L=-1.20\ut{m/s},$$ $$ \begin{aligned} v_R&=-\frac{m_L}{m_R}v_L\\ &=\frac{12}{5}\ut{m/s} \end{aligned} $$ $$ \begin{aligned} S&=vt\\ &=\frac{48}{25}\ut{m}\\ &=1.92\ut{m} \end{aligned} $$ $$\ab{b}$$ $$v_{L\larr R}=-1.20\ut{m/s},$$ $$ \begin{cases} m_Lv_L+m_Rv_R=0\\ ..

$$ \begin{cases} m_\alpha&=4.00\ut{u}\\ m_{\text{O}}&=16.0\ut{u}\\ \theta_\alpha&=-64.0\degree\\ v_{\text{O}f}&=1.20\times10^5\ut{m/s}\\ \theta_{\text{O}}&=40.0\degree\\ \end{cases} $$ $$\Delta \Sigma \vec p=0,$$ $$ \begin{cases} p_{xi}&=\Sigma p_{xf}\\ 0&=\Sigma p_{yf}\\ \end{cases} $$ $$ \begin{cases} m_\alpha v_{\alpha i}&=m_\alpha v_{\alpha x f}+m_{\text{O}} v_{\text{O} x f}\\ 0&=m_\alpha v_..

$$ \begin{cases} \vec r(t\ut{s})&=(3500-160t)\i+2700\j+300\k\ut{m}\\ m&=250\ut{kg} \end{cases} $$ $$ \begin{aligned} \vec v&=\dyt{\vec r}\\ &=-160\i\ut{m/s} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \vec p&=m\vec v\\ &=-40000\i\ut{kg\cdot m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec v&=-160\i\ut{m/s}, \end{aligned} $$ $$\text{West}$$ $$\ab{c}$$ $$ \begin{aligned} \vec F&=m\vec..

$$ \begin{cases} h&=4.0\ut{m}\\ \Delta t&=1.5\ut{ms}=1.5\times10^{-3}\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2(-g)(-h)&=v^2-{0}^2\\ v&=\sqrt{2gh}\\ \end{aligned} $$ $$ \begin{aligned} \abs{F}&=\abs{\frac{\Delta \vec p}{\Delta t}}\\ \abs{s(mg)}&=\abs{\frac{m(\vec v_f-\vec v_i)}{\Delta t}}\\ smg&=\frac{m\sqrt{2gh}}{\Delta t}\\ \end{aligned} $$ $$ \begin..

$$ \begin{cases} m_A&=1.0\ut{kg}\\ m_B&=3.0\ut{kg}\\ v_A&=2.5\ut{m/s}\\ v_\com&=0\\ \end{cases} $$ $$r_\com=\frac{\Sigma rm}{M},$$ $$ \begin{aligned} v_\com&=\dyt{r_\com}\\ &=\dt\(\frac{\Sigma rm}{M}\)\\ &=\frac{1}{M}\sum\dyt{rm}\\ &=\frac{1}{M}\sum \(m\dyt{r}\)\\ &=\frac{1}{M}\sum \(mv\)\\ &=\frac{\Sigma mv}{M}\\ 0&=\frac{m_Av_A+m_Bv_B}{M}\\ \end{aligned} $$ $$ \begin{aligned} \abs{v_B}&=\abs{-..

$$ \begin{cases} m_A&=1200\ut{kg}\\ m_B&=1800\ut{kg}\\ \vec J &= 300\ut{N\cdot s}\\ \end{cases} $$ $$ \begin{cases} \Delta \Sigma \vec p=0\\ \vec J =\Delta \vec p \end{cases} $$ $$ \begin{cases} m_B\vec v_B-m_A\vec v_A&=\vec J\\ m_B\vec v_B+m_A\vec v_A&=0 \end{cases} $$ $$ \begin{cases} \vec v_A&=-\frac{\vec J}{2m_A}\\ \vec v_B&=\frac{\vec J}{2m_B}\\ \end{cases} $$ $$ \begin{aligned} \Ans &= \ve..