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)

60 unit

Answer 2

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