9-26 할리데이 11판 솔루션 일반물리학

$$\begin{cases} m_A&=2.00\times10^{-3}\ut{kg}\\ \vec F_A&=4.00\i+5.00\j\ut{N}\\ m_B&=4.00\times 10^{-3}\ut{kg}\\ \vec F_B&=2.00\i-4.00\j\ut{N}\\ v_0&=0\\ t&=2.00\ut{ms}=2\times10^{-3}\ut{s} \end{cases}$$ $$r_\com=\frac{\Sigma rm}{M},$$ $$\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}\\ &=\frac{1}{M}\sum (m\vec a)\\ &=\frac{\Sigma \vec F}{M}\\ &=\frac{\vec F_A+\vec F_B}{m_A+m_B}\\ &=10^3\i+\frac{500}{3}\j\ut{m/s^2} \end{aligned}$$ $$\begin{aligned} \vec v_\com&=\vec v_0+\Delta \vec v\\ &=\vec v_0+\int_0^t \vec a \dd t\\ &=\vec a t\\ \end{aligned}$$ $$\begin{aligned} \Delta \vec r_\com&=\int_0^t \vec v\dd t\\ &=\int_0^t (\vec a t)\dd t\\ &=\frac{1}{2}\vec a t^2\\ &=\(2\i+\frac{1}{3}\j\)\times10^{-3}\ut{m}\\ &=2\i+\frac{1}{3}\j\ut{mm}\\ \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} \Delta r_\com&=\sqrt{2^2+\(\frac{1}{3}\)^2}\\ &=\frac{\sqrt{37}}{3}\ut{mm}\\ &\approx 2.0275875100994063\ut{mm}\\ &\approx 2.03\ut{mm}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \theta&=\tan^{-1}\frac{\frac{1}{3}}{2}\\ &\approx 0.16514867741462683\ut{rad}\\ &\approx 0.165\ut{rad}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \KE&=\frac{1}{2}mv^2\\ &=\frac{1}{2}M(at)^2\\ &=\frac{1}{2}Ma^2t^2\\ &=\frac{37}{3000}\ut{J}\\ &\approx 0.012333333333333333\ut{J}\\ &\approx 1.23\times10^{-2}\ut{J}\\ &\approx 12.3\ut{mJ}\\ \end{aligned}$$