Question

A small particle of mass $m$ is projected at an angle $\theta$ with the $x$ axis with an initial velocity $v_0$ in the $xy$ plane as shown in the figure. At a time $t<\frac{v_0\sin \theta }{g}$, the angular momentum of the particle is:
(here $\hat{i},\hat{j}$ and $\hat{k}$ are unit vectors along $x,y$ and $z$-axis respectively)

• A. $-mg{v}_0{t}^2\cos \theta \hat{j}$
• B. $mg{v}_{0}t\cos \theta \hat{k}$
• C. $-\frac{1}{2}mg{v}_0{t}^2\cos \theta \hat{k}$
• D. $\frac{1}{2}mg{v}_0{t}^2\cos \theta \hat{i}$

Answer 1

Answer: (C)

$-\frac{1}{2}mg{v}_0{t}^2\cos \theta \hat{k}$

$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
$\overrightarrow{L}=m[v_0\cos \theta t\hat{i}+(v_0\sin \theta t-\frac{1}{2}g{t}^2\left)\hat{j}\right]\times [v_0\cos \theta \hat{i}+(v_0\sin \theta -gt\left)\hat{j}\right]$
$=m{v}_0\cos \theta t[-\frac{1}{2}gt]\hat{k}$
$=-\frac{1}{2}mg{v}_0{t}^2\cos \theta \hat{k}$