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)

60 units

Answer 2

$\overrightarrow{L}=\overrightarrow{r} \times \overrightarrow{p}$
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
$\Rightarrow \, \, OR=OPsin45^{\circ}$
$=4\frac{1}{\sqrt 2}=\frac{4}{\sqrt 2}=2\sqrt 2$
Angular momentum of particle going along this line
$=r \times mv=2\sqrt 2 \times 5 \times 3\sqrt 2 =60$ units