11판/9. 질량중심과 선운동량

9-19 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 17. 21:28
$$ \begin{cases} m_A&=m_B=m=500\ut{g}\\ x_A&=0\\ x_B&=50\ut{mm}\\ \end{cases} $$ $$r_\com=\frac{\Sigma rm}{M},$$ $$ \begin{aligned} x_\com&=\frac{\Sigma xm}{M}\\ &=\frac{x_Am_A+x_Bm_B}{2m}\\ &=\frac{m_B}{2m}x_B\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} x_\com&=\frac{m_B}{2m}x_B\\ &=\frac{m}{2m}x_B\\ &=\frac{x_B}{2}\\ &=25\ut{mm} \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} m_A&=m-\Delta m\\ m_B&=m+\Delta m\\ \Delta m&=20\ut{g}\\ \end{cases} $$ $$ \begin{aligned} x_\com&=\frac{m_B}{2m}x_B\\ &=\frac{m+\Delta m}{2m}x_B\\ &=26\ut{mm}\\ \end{aligned} $$ $$\ab{c,d}$$ $$ \begin{cases} m_A&=m-\Delta m\\ m_B&=m+\Delta m\\ \Delta m&=20\ut{g}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_A&=m_Aa_A\\ \Sigma F_B&=m_Ba_B\\ a_A&=-a_B=a \end{cases} $$ $$ \begin{cases} T-m_Ag&=m_Aa\\ T-m_Bg&=m_B(-a)\\ \end{cases} $$ $$ \begin{aligned} a&=\frac{m_B-m_A}{M}g\\ &=\frac{(m+\Delta m)-(m-\Delta m)}{2m}g\\ &=\frac{\Delta m}{m}g\\ \end{aligned} $$ $$ \begin{aligned} a_\com&=\dytt{r_\com}\\ &=\dtt\(\frac{\Sigma rm}{M}\)\\ &=\frac{1}{M}\sum \(\dytt{rm}\)\\ &=\frac{1}{M}\sum \(m\dytt{r}\)\\ &=\frac{1}{M}\sum \(ma\)\\ &=\frac{\Sigma ma}{M}\\ &=\frac{m_Aa_A+m_Ba_B}{M}\\ &=\frac{m_Aa+m_B(-a)}{M}\\ &=\frac{\bra{(m-\Delta m)-(m+\Delta m)}a}{2m}\\ &=\frac{(-2\Delta m)}{2m}\cdot\frac{\Delta m}{m}g\\ &=-\(\frac{\Delta m}{m}\)^2g\\ &=-\frac{g}{625}\\ &=-0.01569064\ut{m/s^2}\\ &= -15.69064\ut{mm/s^2}\\ &\approx -16\ut{mm/s^2}\\ \end{aligned} $$ $$\ab{c}$$ $$\text{Down}$$ $$\ab{d}$$ $$-16\ut{mm/s^2}$$