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 :
Updated On: Jul 12, 2022
60 unit
$ 40\sqrt{2} $ unit
zero
7.5 unit
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The Correct Option isA
Solution and Explanation
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
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