A body of mass 5 ,kg moving with a uniform sp | vedclass

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(B) The equation of the line is $x - y + 4 = 0$.The angular momentum $L$ of a particle about the origin is given by $L = mvd$,where $m$ is the mass,$v

Vedclass · Accepted Answer

$60$

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(B) The equation of the line is $x - y + 4 = 0$.The angular momentum $L$ of a particle about the origin is given by $L = mvd$,where $m$ is the mass,$v$ is the speed,and $d$ is the perpendicular distance from the origin to the line of motion.The perpendicular distance $d$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$A = 1$,$B = -1$,and $C = 4$. So,$d = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \,m$.Given mass $m = 5 \,kg$ and speed $v = 3\sqrt{2} \,m/s$.Therefore,$L = 5 	imes (3\sqrt{2}) 	imes (2\sqrt{2}) = 5 	imes 3 	imes 2 	imes 2 = 60 \,kg \,m^2/s$.

$A$ body of mass $5 \,kg$ moving with a uniform speed $3 \sqrt{2} \,m/s$ in the $X-Y$ plane along the line $y=x+4$. The angular momentum of the particle about the origin will be . . . . . . $kg \,m^2/s$.

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