Question

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

• A. $60$ units
• B. $40\sqrt{2}$ units
• C. zero
• D. $7.5$ units

Answer 1

Answer: (A)

$\vec{L}=\vec{r}\times \vec{p}$
$Y+X+4$ line has been shown in the figure.

$Y=4$, So $OP=4$,
The slope of the line can be obtained by comparing with the equation of line
$y=mx+c$
$m=\tan \theta =1$
$\Rightarrow \theta =45^{\circ }$
$\angle OQP=\angle OPQ=45^{\circ }$
If we draw a line perpendicular to this line.
Length of the perpendicular $=OR$
$\Rightarrow OR=OP\sin 45^{\circ }$
$=4\frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$
Angular momentum of particle going along this line
$=r\times mv=2\sqrt{2}\times 5\times 3\sqrt{2}=60$ units