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Figure 7.4 shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines.
At a particular instant, $r_{1}$ and $r_{2}$ are their respective position vectors drawn from point A which is in the plane of the parallel lines . Choose the correct options :

(a) Angular momentum $I_{1}$ of particle 1 about A is $I_{1}=mvd_{1} \odot$
(b) Angular momentum $I_{2}$ of particle 2 about A is $I_{2}=mvr_{2}\odot$
(c) Total angular momentum of the system about A is $I=mv(r_{1}+r_{2})\odot$
(d) Total angular momentum of the system about A is $I=mv(d_{2}-d_{1})\otimes$
$\odot$ represents a unit vector coming out of the page.
$\otimes$ represents a unit vector going into the page.

Answers (1)

The correct answer is the option (a) and (d)

Using $\overrightarrow{L}=\vec{r}\times \vec{p}$ the direction of L is determined as perpendicular to the plane of r and p by Right-hand thumb rule.

$\overrightarrow{L_{1}}=\vec{r}\times m\vec{v}$ (out of the page) $=m\vec{v}d_{1}$

$\overrightarrow{L_{2}}=\overrightarrow{r_{2}}\times m\overrightarrow{(-v)}$ (into the page) $=m\vec{v}d_{2}$

Hence a is correct and b wrong.

For total angular momentum $\vec{L}=\overrightarrow{L_{1}}+\overrightarrow{L_{2}}$

$\vec{L}=m\vec{v}d_{1}$ (out of page) $-m\vec{v}d_{2}$ (into page)

$\vec{L}=m\vec{v}(-d_{2}-d_{1})$ [ into the page]

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