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Q. A particle of mass 5g is moving with a speed of $3\sqrt2cms^{-1}$ in X-Y plane along the line y = x + 4 . The magnitude of its angular momentum about the origin in $gcm^2s^{-1}$ is

Solution:

Given $y = x+4$ when $x = 0, y = 4cm$ and $y = 0,x = - 4cm$ perpendicular distance $OC = OA $ sin $45^\circ = 4 \times \frac{1}{\sqrt{2}} = 2 \sqrt{2} cm$ Angular momentum = $m \nu (OC ) = 5 ( 3 \sqrt{2} ) ( 2 \sqrt{2} )$ = $60 \, gm \, cm^2 \, s^{-1}$