Question

A particle of mass $m=5$ units is moving with a uniform speed $v=3\sqrt{2}$ units 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

Answer: (A)

Momentum of the particle
=mass $\times$ velocity
$=(5)\times (3\sqrt{2})=15\sqrt{2}$
The direction of momentum in the XOY plane is given by
$y=x+4$
Slope of the line $=1=\tan \theta$

i.e., $\theta =45^{\circ }$.
Intercept of its straight line $=4$
Length of the perpendicular $z$ from the origin of the straight line
$=4\sin 45^{\circ }=\frac{4}{\sqrt{2}}=2\sqrt{2}$
Angular momentum = momentum $\times$ perpendicular length
$=15\sqrt{2}\times 2\sqrt{2}=60$ unit