솔루션피아

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

$$ \begin{cases} m&=0.15\ut{kg}\\ v_i&=5.00\i+6.50\j+4.00\k\ut{m/s}\\ v_f&=2.00\i+3.50\j-3.20\k\ut{m/s}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Delta \vec p&=\Delta (m\vec v)\\ &=m\Delta \vec v\\ &=-\frac{9}{20}\i-\frac{9}{20}\j-\frac{27}{25}\k\ut{kg\cdot m/s}\\ &=-0.45\i-0.45\j-1.08\k\ut{kg\cdot m/s}\\ &\approx -0.45\i-0.45\j-1.1\k\ut{kg\cdot m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin..

(풀이자 주 : 당구공의 질량에 대한 언급이 없습니다. 풀이자 임의로 두 당구공의 질량이 같다고 두겠습니다.) $$ \begin{cases} m_1&=m_2=m\\ v_{1i}&=2.2\ut{m/s}\\ v_{2i}&=0\\ v_{2f}&=1.1\ut{m/s}, \theta_2=60\degree\\ \end{cases} $$ $$ \begin{cases} \vec v_{1i}&=v_{1i}\i\\ \vec v_{2f}&=v_{2f}\cos\theta_2\i+v_{2f}\sin\theta_2\j \end{cases} $$ $$\Delta \Sigma \vec P=0,$$ $$m\vec v_{1i}=m\vec v_{1f}+m\vec v_{2f},$$ $$\vec v_{1i}=\vec v_{1f}+\vec v..

$$ \put\begin{cases} 0:\text{Start}\\ 1:\text{Highest Point Before Bomb}\\ 2:\text{Highest Point After Bomb}\\ 3:\text{Landing}\\ \end{cases} $$ $$ \begin{cases} v_0&=24\ut{m/s}\\ \theta_0&=60\degree\\ v_{A1}&=0 \end{cases} $$ $$2aS=v^2-{v_0}^2,$$ $$2(-g)(h)=(0)^2-{v_{y0}}^2$$ $$ \begin{aligned} h_1&=\frac{(v_0\sin\theta)^2}{2g}\taag1\\ \end{aligned} $$ $$ \begin{aligned} v_1&=v_x\\ &=v_0\cos\th..

$$ \begin{cases} m_1&=200\ut{g}=0.2\ut{kg}\\ v_{1i}&=3.00\ut{m/s}\\ m_2&=400\ut{g}=0.4\ut{kg}\\ v_{2i}&=0\\ \end{cases} $$ $$ \put\begin{cases} \vec v_{\text{in}}&=\vec v_{1i}-\vec v_{2i}=v_{1i}\\ \vec v_{\text{out}}&=\vec v_{1f}-\vec v_{2f}\\ \end{cases} $$ $$\Delta \Sigma \vec P=0,$$ $$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$$ $$ \begin{aligned} J&=\Delta \vec P\\ &=m_1v_{1f}-m_1v_{1i}\\ &=m_1..

$$ \begin{cases} m_0&=2.0\ut{g}=2.0\times10^{-3}\ut{kg}\\ f&=10\ut{Hz}\\ v&=500\ut{m/s}\\ \end{cases} $$ $$\ab{a}$$ $$ p_0 =m_0v=1\ut{kg\cdot m/s} $$ $$\ab{b}$$ $$ \KE =\frac{1}{2}m_0v^2=250\ut{J} $$ $$\ab{c}$$ $$ \begin{aligned} \vec F&=\dyt{\vec p}\\ &=\dyt{m\vec v}\\ &=\vec v\dyt{m_0 N}\\ &=m_0\vec v\dyt{ N}\\ &=p_0f\\ &=10\ut{N} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \vec F &= \frac{..

$$ \begin{cases} m_A&=2.00\times10^{-3}\ut{kg}\\ \vec F_A&=4.00\i+5.00\j\ut{N}\\ m_B&=4.00\times 10^{-3}\ut{kg}\\ \vec F_B&=2.00\i-4.00\j\ut{N}\\ v_0&=0\\ t&=2.00\ut{ms}=2\times10^{-3}\ut{s} \end{cases} $$ $$r_\com=\frac{\Sigma rm}{M},$$ $$ \begin{aligned} \vec a_\com&=\dytt{\vec r_\com}\\ &=\dtt\(\frac{\Sigma \vec rm}{M}\)\\ &=\frac{1}{M}\sum \dytt{\vec rm}\\ &=\frac{1}{M}\sum m\dytt{\vec r}\\ ..

$$ \put \begin{cases} S:\text{Ship}\\ J:\text{Jupiter}\\ \end{cases} $$ $$ \begin{cases} m_S&=m\\ m_J&=M\\ v_{Si}&=10.5\ut{km/s}\\ v_{Ji}&=13.0\ut{km/s}\\ m_J&\gg m_S\\ \end{cases} $$ $$ \begin{cases} \Delta \Sigma \vec P=0,\\ \text{Elastic Collision}\Harr\vec v_{\text{in}}+ \vec v_{\text{out}}=0, \end{cases} $$ $$ \begin{cases} m_Sv_{Si}+m_Jv_{Ji}=m_Sv_{Sf}+m_Jv_{Jf}\\ v_{Si}-v_{Ji}+v_{Sf}-v_{J..