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Physics

Mechanics

Easy

Question

$text A particle of mass end text 1 kg text is moving along the line end text straight y equals straight x plus 2 text (here end text straight x text and end text straight y text are in end text straight m text ) with speed end text 2 straight m divided by straight s text The magnitude of angular momentum of the particle about origin is: end text$

$4 {kgm}^{2} s^{- 1}$
$2 \sqrt{2} {kgm}^{2} s^{- 1}$
$4 \sqrt{2} {kgm}^{2} s^{- 1}$
$2 {kgm}^{2} s^{- 1}$

The correct answer is: $2 square root of 2 kgm squared straight s to the power of negative 1 end exponent$

$L equals m v subscript 1 equals left parenthesis 1 right parenthesis left parenthesis 2 right parenthesis open parentheses 2 cos invisible function application 45 to the power of ring operator close parentheses equals fraction numerator 4 over denominator square root of 2 end fraction equals 2 square root of 2 kgms squared s to the power of negative 1 end exponent$

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