A particle of mass m is moving along a line y | vedclass

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(C) The equation of the line is $y = x + a$. Comparing this with the slope-intercept form $y = mx + c$, we get the slope $m_{slope} = 1$. Since $	an

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$\frac{mva}{\sqrt{2}}$

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(C) The equation of the line is $y = x + a$. Comparing this with the slope-intercept form $y = mx + c$, we get the slope $m_{slope} = 1$. Since $	an 	heta = m_{slope} = 1$, the angle $	heta = 45^{\circ}$.The velocity vector $\vec{v}$ makes an angle of $45^{\circ}$ with the positive $x$-axis. Thus, $\vec{v} = v \cos 45^{\circ} \hat{i} + v \sin 45^{\circ} \hat{j} = \frac{v}{\sqrt{2}} (\hat{i} + \hat{j})$.The angular momentum $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p} = m (\vec{r} 	imes \vec{v})$.The perpendicular distance $d$ from the origin $(0, 0)$ to the line $x - y + a = 0$ is given by $d = \frac{|(1)(0) - (1)(0) + a|}{\sqrt{1^2 + (-1)^2}} = \frac{a}{\sqrt{2}}$.The magnitude of angular momentum is $L = mvd = m v \left( \frac{a}{\sqrt{2}} ight) = \frac{mva}{\sqrt{2}}$.

$A$ particle of mass $m$ is moving along a line $y = x + a$ with a constant velocity $v$. The angular momentum of the particle about the origin is

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