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(D) Let the intersection point $O$ be the origin $(0,0)$. Let one particle $Q$ move along the $x$-axis towards $O$ and the other particle $P$ move alo

Vedclass · Accepted Answer

$50\sqrt{3} \, cm$

Vedclass · Answer

(D) Let the intersection point $O$ be the origin $(0,0)$. Let one particle $Q$ move along the $x$-axis towards $O$ and the other particle $P$ move along a line at $60^{\circ}$ to the $x$-axis towards $O$.The velocities are $\vec{v}_Q = -20 \hat{i} \, cm/s$ and $\vec{v}_P = -20 \cos 60^{\circ} \hat{i} - 20 \sin 60^{\circ} \hat{j} = -10 \hat{i} - 10\sqrt{3} \hat{j} \, cm/s$.The relative velocity of $Q$ with respect to $P$ is $\vec{v}_{QP} = \vec{v}_Q - \vec{v}_P = (-20 - (-10)) \hat{i} - (-10\sqrt{3}) \hat{j} = -10 \hat{i} + 10\sqrt{3} \hat{j} \, cm/s$.The initial positions are $\vec{r}_Q = 400 \hat{i} \, cm$ and $\vec{r}_P = 300 \cos 60^{\circ} \hat{i} + 300 \sin 60^{\circ} \hat{j} = 150 \hat{i} + 150\sqrt{3} \hat{j} \, cm$.The relative position is $\vec{r}_{QP} = \vec{r}_Q - \vec{r}_P = (400 - 150) \hat{i} - 150\sqrt{3} \hat{j} = 250 \hat{i} - 150\sqrt{3} \hat{j} \, cm$.The shortest distance is given by $d = \frac{|\vec{r}_{QP} 	imes \vec{v}_{QP}|}{|\vec{v}_{QP}|}$.$|\vec{v}_{QP}| = \sqrt{(-10)^2 + (10\sqrt{3})^2} = \sqrt{100 + 300} = 20 \, cm/s$.The cross product $\vec{r}_{QP} 	imes \vec{v}_{QP} = (250 \hat{i} - 150\sqrt{3} \hat{j}) 	imes (-10 \hat{i} + 10\sqrt{3} \hat{j}) = (250 	imes 10\sqrt{3} - (-150\sqrt{3}) 	imes (-10)) \hat{k} = (2500\sqrt{3} - 1500\sqrt{3}) \hat{k} = 1000\sqrt{3} \hat{k}$.The magnitude is $1000\sqrt{3}$.Therefore,$d = \frac{1000\sqrt{3}}{20} = 50\sqrt{3} \, cm$.

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