Problem: (2 Parts) A ball of mass, $m=2.0 \mathrm{~kg}$ is falling straight down. At the instant shown, it is at height, $h$ $=7.0 \mathrm{~m}$ above the ground and has vertically downward speed, $\mathrm{v}=6.0 \mathrm{~m} / \mathrm{s}$. $6 p t$ Part A: At that instant of time, what is the translational angular momentum, $\vec{L}_{\text {trans, } P}$ of the ball relative to the point $\mathrm{P}$ with position vector, $\overrightarrow{\mathrm{r}}_{\mathrm{P}}=\langle-3.0,0,0\rangle \mathrm{m}$ as shown in the figure. \[ \begin{aligned} \text { 11.A } \bigcirc & <0,0,-36>k g \cdot \mathrm{m}^{2} / \mathrm{s} \\ \mathbf{B} \bigcirc & <0,0,36>\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} \\ \mathbf{C} \bigcirc & <0,0,-84>\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} \\ \mathbf{D} \bigcirc & <0,0,-91>\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} \end{aligned} \] EO None of the above
Part B: At the instant considered in the preceding problem what is, $\frac{d \vec{L}_{\text {trans, } P}}{d t}$, the rate of change of the ball's translational angular momentum about point $\mathbf{P}$ ? [Use acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ] 12.A $\bigcirc<0,0,60>\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}$ \[ \begin{array}{l} \mathbf{B} \bigcirc<0,0,-60>\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} \\ \mathbf{C} \bigcirc<0,0,-152>\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} \\ \mathbf{D} \bigcirc<0,0,-140>\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} \end{array} \] E○ None of the above

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ZM3ZZD The First Answerer