9-33 할리데이 11판 솔루션 일반물리학

$$\begin{cases} m_A&=1.0\ut{kg}\\ m_B&=3.0\ut{kg}\\ v_A&=2.5\ut{m/s}\\ v_\com&=0\\ \end{cases}$$ $$r_\com=\frac{\Sigma rm}{M},$$ $$\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{-\frac{m_A}{m_B}v_A}\\ &=\frac{5}{6}\ut{m/s}\\ &\approx 0.8333333333333334\ut{m/s}\\ &\approx 0.83\ut{m/s}\\ &\approx 83\ut{cm/s} \end{aligned}$$