A particle of mass = 5 text{ units} is moving | vedclass

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(A) The momentum of the particle is given by $p = m 	imes V = 5 	imes 3 \sqrt{2} = 15 \sqrt{2} 	ext{ units}$.The particle moves along the line $Y =

Vedclass · Accepted Answer

$60$

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(A) The momentum of the particle is given by $p = m 	imes V = 5 	imes 3 \sqrt{2} = 15 \sqrt{2} 	ext{ units}$.The particle moves along the line $Y = X + 4$,which can be rewritten as $X - Y + 4 = 0$.The perpendicular distance $r_{\perp}$ from the origin $(0, 0)$ to the line $AX + BY + C = 0$ is given by $r_{\perp} = \frac{|C|}{\sqrt{A^2 + B^2}}$.Substituting the values $A = 1, B = -1, C = 4$,we get $r_{\perp} = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} 	ext{ units}$.The magnitude of the angular momentum $L$ about the origin is $L = p 	imes r_{\perp}$.$L = (15 \sqrt{2}) 	imes (2 \sqrt{2}) = 15 	imes 2 	imes 2 = 60 	ext{ units}$.

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

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