Question

A particle of mass $m=5$ is moving with a uniform speed $v=3 \sqrt{2}$ in the $X O Y$ 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 \quad $
$\Rightarrow \theta=45^{\circ} $
$\angle O Q P=\angle O P Q=45^{\circ}$
If we draw a line perpendicular to this line.
Length of the perpendicular $= OR$
$\Rightarrow O R=O P \sin 45^{\circ}$
$=4 \frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2 \sqrt{2}$
Angular momentum of particle going along this line
$=r \times m v=2 \sqrt{2} \times 5 \times 3 \sqrt{2}=60$ units