11판/9. 질량중심과 선운동량

9-15 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 16. 20:39
put {a:mab:mb \put \begin{cases} a:m_a\\ b:m_b \end{cases} {vai=0.75[m/s]ma=0.25[kg]mb=0.50[kg]La=5.0[cm]=5×102[m]Lb=6.0[cm]=6×102[m]xai=1.50[m]xbi=0[m] \begin{cases} v_{ai}&=0.75\ut{m/s}\\ m_a&=0.25\ut{kg}\\ m_b&=0.50\ut{kg}\\ L_a&=5.0\ut{cm}=5\times10^{-2}\ut{m}\\ L_b&=6.0\ut{cm}=6\times10^{-2}\ut{m}\\ x_{ai}&=-1.50\ut{m}\\ x_{bi}&=0\ut{m}\\ \end{cases} put M=ma+mb=0.75[kg]\put M=m_a+m_b=0.75\ut{kg} rcom=ΣmrM,\vec r_{\com} = \frac{\Sigma m\vec r}{M}, xcom=ΣmxM=maxa+mbxbM \begin{aligned} x_\com&=\frac{\Sigma mx}{M}\\ &=\frac{m_ax_a+m_bx_b}{M} \end{aligned} (a)\ab{a} t=0,t=0, {xa=1.5[m]xb=0 \begin{cases} x_a&=-1.5\ut{m}\\ x_b&=0\\ \end{cases} xcom=maxa+mbxbM=maMxa=12[m]=0.50[m]=50[cm] \begin{aligned} x_\com&=\frac{m_ax_a+m_bx_b}{M}\\ &=\frac{m_a}{M}x_a\\ &=-\frac{1}{2}\ut{m}\\ &=-0.50\ut{m}\\ &=-50\ut{cm}\\ \end{aligned} (b)\ab{b} t=Contact,t=\text{Contact}, {xa=Lb2La2=5.5[cm]xb=0 \begin{cases} x_a&=-\frac{L_b}{2}-\frac{L_a}{2}=-5.5\ut{cm}\\ x_b&=0\\ \end{cases} xcom=maxa+mbxbM=maMxa=116[cm]1.8333333333333333[cm]1.8[cm] \begin{aligned} x_\com&=\frac{m_ax_a+m_bx_b}{M}\\ &=\frac{m_a}{M}x_a\\ &=-\frac{11}{6}\ut{cm}\\ &\approx -1.8333333333333333\ut{cm}\\ &\approx -1.8\ut{cm}\\ \end{aligned} (c)\ab{c} vcom= ⁣dxcom ⁣dt= ⁣d ⁣dtmaxa+mbxbM=ma ⁣dxa ⁣dt+mb ⁣dxb ⁣dtM=mavai+mbvbiM=maMvai \begin{aligned} v_\com&=\dyt{x_\com}\\ &=\dt\frac{m_ax_a+m_bx_b}{M}\\ &=\frac{m_a\dyt{x_a}+m_b\dyt{x_b}}{M}\\ &=\frac{m_av_{ai}+m_bv_{bi}}{M}\\ &=\frac{m_a}{M}v_{ai}\\ \end{aligned} xcom(t)=xcom(0)+vcomt=maMxai+maMvait=maM(xai+vait)=14(t2) \begin{aligned} x_\com(t)&=x_\com(0)+v_\com t\\ &=\frac{m_a}{M}x_{ai}+\frac{m_a}{M}v_{ai}t\\ &=\frac{m_a}{M}(x_{ai}+v_{ai}t)\\ &=\frac{1}{4}(t-2)\\ \end{aligned} xcom(4)=12[m]=0.50[m] \begin{aligned} x_\com(4)&=\frac{1}{2}\ut{m}\\ &=0.50\ut{m}\\ \end{aligned}