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 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. $-mg{v}_0{t}^{2}cos\theta \hat{j}$
• B. $mg{v}_{0}tcos\theta \hat{k}$
• 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

The correct option is C $-\frac{1}{2}mg{v}_0{t}^{2}cos\theta \hat{k}$

As we know that,

angular momentum =$m(\overrightarrow{r}\times \overrightarrow{v})$

where $\overrightarrow{r}$=position vector at time t

$\overrightarrow{v}$=velocity vector at time t

here $\overrightarrow{r}=v_o\cos \theta t\hat{i}+\left(v_o\sin \theta t-\frac{1}{2}g{t}^2\right)\hat{j}$

and, $\overrightarrow{v}=v_o\cos \theta \hat{i}+\left(v_o\sin \theta -gt\right)\hat{j}$

so angular momentum= $\{v_o\cos \theta t\hat{i}+\left(v_o\sin \theta t-\frac{1}{2}g{t}^2\right)\hat{j}\}\times \{v_o\cos \theta \hat{i}+\left(v_o\sin \theta -gt\right)\hat{j}\}$

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