Question

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

• A. $60$ unit
• B. $40\sqrt{2}$ unit
• C. Zero
• D. $7.5$ unit

Answer 1

The correct option is A $60$ unit

Momentum of the particle
$=mass\times velocity=\left(5\right)\times \left(3\sqrt{2}\right)=15\sqrt{2}$
The direction of momentum in the XOY plane is given by
$y=x+4$
Slope of the line $=1=tan\theta$
i.e., $\theta ={45}^o$
Intercept of its straight line=4
Length of the perpendicular z from the origin of the straight line.
$=4sin{45}^o=\frac{4}{\sqrt{2}}=2\sqrt{2}$
Angular momentum,
$=momentum\times perpendicularlength$
$=15\sqrt{2}\times 2\sqrt{2}=60$ units