If a particle of mass $m$ is moving with constant velocity $v$ parallel to $X-$axis in $x-y$ plane as shown in fig. Its angular momentum with respect to origin at any time $t$ will be
$mvb\,\hat k$
$ - mvb\, \,\hat k$
$mvb\,\hat i$
$mv\,\hat i$
If a particle of mass $m$ is moving with constant velocity $v$ parallel to $X-$axis in $x-y$ plane as shown in fig. Its angular momentum with respect to origin at any time $t$ will be
We know that, Angular momentum
$\overrightarrow L = \overrightarrow {r\,} \times \overrightarrow p $ in terms of omponent becomes
$\overrightarrow L = \left| {\,\begin{array}{*{20}{c}}
{\hat i\,\,}&{\hat j\,\,}&{\hat k} \\
{x\,\,}&{y\,\,}&z \\
{{p_x}}&{\,\,{p_y}\,\,}&{{p_z}}
\end{array}\,} \right|$
As motion is in x-y plane ($z = 0 $ and ${P_z} = 0$), so $\overrightarrow {L\,} = \overrightarrow {k\,} (x{p_y} - y{p_x})$
Here $x = vt, y = b, $${p_x} = m\,v$ and ${p_y} = 0$
$\therefore \overrightarrow {L\,} = \overrightarrow {k\,} \left[ {vt \times 0 - b\,mv} \right] = - mvb\,\hat k$
$\overrightarrow L = \overrightarrow {r\,} \times \overrightarrow p $ in terms of omponent becomes
$\overrightarrow L = \left| {\,\begin{array}{*{20}{c}}
{\hat i\,\,}&{\hat j\,\,}&{\hat k} \\
{x\,\,}&{y\,\,}&z \\
{{p_x}}&{\,\,{p_y}\,\,}&{{p_z}}
\end{array}\,} \right|$
As motion is in x-y plane ($z = 0 $ and ${P_z} = 0$), so $\overrightarrow {L\,} = \overrightarrow {k\,} (x{p_y} - y{p_x})$
Here $x = vt, y = b, $${p_x} = m\,v$ and ${p_y} = 0$
$\therefore \overrightarrow {L\,} = \overrightarrow {k\,} \left[ {vt \times 0 - b\,mv} \right] = - mvb\,\hat k$