Q.
A small particle of mass $m\ $is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_{0}$ in the $x$-$y$ plane as shown in the figure. $At\ a\ time\ t < \frac{v_{0}\sin\theta}{g},$ the angular momentum of the particle is
A) $- mgv_{0}t^{2}\cos{\theta\ \widehat{\dot{\mathbf{j}}}}$
B) $mgv_{0}t\cos{\theta\ \widehat{\mathbf{k}}}\ $
C) $- \frac{1}{2}mgv_{0}t^{2}\cos{\theta\ \widehat{\mathbf{k}}}$
D) $\frac{1}{2}mgv_{0}t^{2}\cos{\theta\ \widehat{\dot{\mathbf{i}}}}$
Explanation
(c) $\mathbf{L} = m\left( \mathbf{r} \times \mathbf{v} \right)$
$\mathbf{L} = m\left\lbrack v_{0}cos\theta\ t\widehat{\dot{\mathbf{i}}} + \left( v_{0}sin\theta\ t - \frac{1}{2}gt^{2} \right)\widehat{\dot{\mathbf{j}}} \right\rbrack$
$\times \left\lbrack v_{0}cos\theta\ \widehat{\dot{\mathbf{i}}} + \left( v_{0}sin\theta - gt \right)\widehat{\dot{\mathbf{j}}} \right\rbrack$
$= mv_{0}cos\theta\ t\left\lbrack - \frac{1}{2}gt \right\rbrack\widehat{\mathbf{k}}$
$= - \frac{1}{2}mgv_{0}t^{2}cos\theta\ \widehat{\mathbf{k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.