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physics

mechanics

Easy

Question

$text For a particle of a mass end text 100 gm text , position and velocity at any instant are given as end text 10 i with not stretchy hat on top plus 6 j with not stretchy hat on top cm text and end text straight v with not stretchy rightwards arrow on top equals 5 straight i with not stretchy hat on top cm divided by straight s text . Calculate the angular momentum about the point end text left parenthesis 1 comma 1 right parenthesis cm.$

$25 \times 10^{- 5} (- \hat{k}) {kgm}^{2} s^{- 1}$
$30 \times 10^{- 5} (- \hat{k}) {kgm}^{2} s^{- 1}$
$3 \times 10^{- 3} (- \hat{k}) {kgm}^{2} s^{- 1}$
$25 \times 10^{- 4} (- \hat{k}) {kgm}^{2} s^{- 1}$

The correct answer is: $25 \times 10^{- 5} (- \hat{k}) {kgm}^{2} s^{- 1}$

$L with not stretchy rightwards arrow on top equals m left parenthesis r with not stretchy rightwards arrow on top cross times c with not stretchy rightwards arrow on top right parenthesis equals 100 over 1000 open square brackets fraction numerator 9 i with not stretchy hat on top plus 5 j with not stretchy hat on top over denominator 100 end fraction close square brackets cross times fraction numerator 5 i with not stretchy hat on top over denominator 100 end fraction equals 25 cross times 10 to the power of negative 5 end exponent left parenthesis negative k with not stretchy hat on top right parenthesis khm squared straight s to the power of negative 2 end exponent$

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