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A particle is moving along a line $y=x+a \mid$ with a constant velocity $v$. Find the angular momentum of the particle about the origin.
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The correct answer is:
$\frac{m v a}{\sqrt{2}}$
$\angle P O Q=45^{\circ}$ So, $O Q=a \cos 45^{\circ}=a / \sqrt{2}$ $L=m v(O Q)=m v a / \sqrt{2}$
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