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The time dependence of the position of a particle of mass $m = 2$ is given by $\vec r\,(t)\, = \,2t\,\hat i\, - 3{t^2}\hat j$ Its angular momentum with respect to the origin at time $t = 2$ is
$-48\,\hat k$
$48\,(\hat i\, + \,\hat j)$
$36\,\hat k$
$ - \,34\,(\hat k\, - \,\hat i)$
Solution
$\overrightarrow v = 2\hat i – 6 + \hat j$
$At\,t = 2$
$\overrightarrow v = 2\hat i – 12\hat j$
$\overrightarrow p = m\overrightarrow v = 4i – 24\hat j$
$At\,t = 2$
$\overrightarrow r = 4\hat i – 12\hat j$
$\overrightarrow L = \overrightarrow r \times \overrightarrow p = \left| \begin{array}{l}
\hat i\,\,\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\,\,\,\hat k\\
4\,\,\,\,\,\, – 12\,\,\,\,\,\,\,\,\,0\\
4\,\,\,\,\,\, – 24\,\,\,\,\,\,\,\,\,0
\end{array} \right|$
$ = \left\{ {4\left( { – 24} \right) + 4 \times 12} \right\}\hat k$
$ = \left( { – 96 + 48} \right)\hat k$
$ = \left( – \right)48\,\hat k$
