A particle of mass m = 5 units is moving in t | vedclass

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(B) The angular momentum $L$ of a particle about the origin is given by $L = p 	imes d$,where $p = mv$ is the linear momentum and $d$ is the perpendi

Vedclass · Accepted Answer

$60$

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(B) The angular momentum $L$ of a particle about the origin is given by $L = p 	imes d$,where $p = mv$ is the linear momentum and $d$ is the perpendicular distance from the origin to the line of motion.Given the line of motion is $Y = X + 4$,which can be written as $X - Y + 4 = 0$.The perpendicular distance $d$ from the origin $(0, 0)$ to the line $X - Y + 4 = 0$ is calculated as:$d = \frac{|(1)(0) - (1)(0) + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.The linear momentum $p = mv = 5 	imes 3\sqrt{2} = 15\sqrt{2}$ units.Therefore,the magnitude of angular momentum is:$L = p 	imes d = (15\sqrt{2}) 	imes (2\sqrt{2}) = 15 	imes 2 	imes 2 = 60$ units.

$A$ particle of mass $m = 5$ units is moving in the $XOY$ plane along the line $Y = X + 4$ with a uniform speed $v = 3\sqrt{2}$ units. The magnitude of its angular momentum about the origin is ...... units.

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