The position vectors of two 1 kg particles, $(A)$ and $(B)$ are given by $\vec{r}_{A}=\left(\alpha_{

Question

The position vectors of two 1 kg particles, $(A)$ and $(B)$ are given by $\vec{r}_{A}=\left(\alpha_{1} t^{2} \hat{i}+\alpha_{2} t \hat{j}+\alpha_{3} t \hat{k}ight) \mathrm{m}$ and $\vec{r}_{B}=\left(\beta_{1} t \hat{i}+\beta_{2} t^{2} \hat{j}+\beta_{3} t \hat{k}ight) \mathrm{m}$, respectively; ( $\alpha_{1}=1 \mathrm{~m} / \mathrm{s}^{2}, \alpha_{2}=3 n \mathrm{~m} / \mathrm{s}, \alpha_{3}=2 \mathrm{~m} / \mathrm{s}, \beta_{1}=2 \mathrm{~m} / \mathrm{s}$, $\beta_{2}=-1 \mathrm{~m} / \mathrm{s}^{2}, \beta_{3}=4 p \mathrm{~m} / \mathrm{s}$ ), where $t$ is time, $n$ and $p$ are constants. At $t=1 \mathrm{~s},\left|\vec{V}_{A}ight|=\left|\vec{V}_{B}ight|$ and velocities $\vec{V}_{A}$ and $\vec{V}_{B}$ of the particles are orthogonal to each other. At $t=1 \mathrm{~s}$, the magnitude of angular momentum of particle $(A)$ with respect to the position of particle $(B)$ is $\sqrt{L} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$. The value of $L$ is $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(90): As, $\vec{r}_{A}=\left(\alpha_{1} t^{2} \hat{i}+\alpha_{2} t \hat{j}+\alpha_{3} t \hat{k}ight) \mathrm{m}$\ $\vec{r}_{B}=\left(\beta_{1} t \hat{i}+\beta_{2} t^{2} \hat{j}+\beta_{3} t \hat{k}ight) \mathrm{m}$\ $\alpha_{1}=1 \mathrm{~m} / \mathrm{s}^{2}, \alpha_{2}=3 n \mathrm{~m} / \mathrm{s}, \alpha_{3}=2 \mathrm{~m} / \mathrm{s}$\ $\beta_{1}=2 \mathrm{~m} / \mathrm{s}, \beta_{2}=-1 \mathrm{~m} / \mathrm{s}^{2}, \beta_{3}=4 \mathrm{pm} / \mathrm{s}$\ $t=1 \mathrm{~s},\left|\vec{V}_{A}ight|=\left|\vec{V}_{B}ight|$\ $\vec{V}_{A}=\frac{d \vec{r}_{A}}{d t}=\left(2 \alpha_{1} t \hat{i}+\alpha_{2} \hat{j}+\alpha_{3} \hat{k}ight)=(2 t \hat{i}+3 n \hat{j}+2 \hat{k})$\ $\vec{V}_{B}=\frac{d \vec{r}_{B}}{d t}=\left(\beta_{1} \hat{i}+2 \beta_{2} t \hat{j}+\beta_{3} \hat{k}ight)=2 \hat{i}-2 t \hat{j}+4 p \hat{k}$\ $\vec{V}_{A} \cdot \vec{V}_{B}=0 ;$ 
$ \begin{aligned} (2 t \hat{i}+3 n \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-2 t \hat{j}+4 p \hat{k}) & =0 \ \Rightarrow(2 \hat{i}+3 n \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+4 p \hat{k}) & =0 \ & (\because t=1 \mathrm{~s}) \end{aligned} $