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(A) The angular momentum $\vec{L}$ is defined as $\vec{L} = \vec{r} 	imes \vec{p}$. Since the particle moves in the $xy$-plane,$\vec{L}$ is directed

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+ ve $x$ axis

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(A) The angular momentum $\vec{L}$ is defined as $\vec{L} = \vec{r} 	imes \vec{p}$. Since the particle moves in the $xy$-plane,$\vec{L}$ is directed along the $z$-axis (perpendicular to the plane of motion).At point $A$,the particle is moving along the $y$-direction (tangent to the ellipse at the extreme right point),so the linear momentum $\vec{p}$ is along the $+y$ direction.We need to find the direction of $\vec{\alpha} = \vec{p} 	imes \vec{L}$.Using the right-hand rule for the cross product:$\vec{p}$ is in the $+y$ direction $(\hat{j})$.$\vec{L}$ is in the $+z$ direction $(\hat{k})$.Therefore,$\vec{\alpha} = \vec{p} 	imes \vec{L} = (p\hat{j}) 	imes (L\hat{k}) = pL(\hat{j} 	imes \hat{k}) = pL\hat{i}$.This corresponds to the $+x$ axis.

$A$ particle is moving in an elliptical orbit as shown in the figure. If $\vec{p}$,$\vec{L}$,and $\vec{r}$ denote the linear momentum,angular momentum,and position vector of the particle (from focus $O$) respectively at point $A$,then the direction of $\vec{\alpha} = \vec{p} \times \vec{L}$ is along

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