Question

A particle is projected at $t=0$ from a point $P$ on the ground with a speed $V_0$, at an angle of ${45}^{\circ }$ to the horizontal. What is the magnitude of the angular momentum of the particle about $P$ at time $t=v_0/g.$

• A. $\frac{m{v}_0^2}{2\sqrt{2}g}$
• B. $\frac{m{v}_0^3}{\sqrt{2}g}$
• C. $\frac{m{v}_0^2}{\sqrt{2}g}$
• D. $\frac{m{v}_0^3}{2\sqrt{2}g}$

Answer 1

The correct option is D $\frac{m{v}_0^3}{2\sqrt{2}g}$

Component of velocity along verticle,$U_x=V_o\cos {45}^0=\frac{v_o}{\sqrt{2}}$

and along horizontal $V_o\cos {45}^o=\frac{V_o}{\sqrt{2}}$

after time $t=\frac{V_o}{g}$ the verticle component of velocity will be

$=\frac{V_o}{\sqrt{2}}-gt$

$=\frac{V_o}{\sqrt{2}}-V_o=V_o\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)$

The horizontal component remains same, so the velocity vector after time $t$

$\overrightarrow{V}=\frac{V_o}{\sqrt{2}}\hat{i}+\frac{V_o}{\sqrt{2}}(1-\sqrt{2})\hat{j}$

In time $t$ particle moves in horizontal direction by a distance.

$x=U_{x}t=\frac{V_o}{\sqrt{2}}\frac{V_o}{g}=\frac{V_o^2}{\sqrt{2}g}$

and a distance along verticle direction

$y=U_{g}t-\frac{1}{2}g{t}^2$

$\frac{V_o}{\sqrt{2}}.\frac{V_o}{g}-\frac{1}{2}g\times \frac{V_o^2}{g^2}$

$=\frac{V_o^2}{2g}\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]$

position vector of body with respect the point of the motion is,

$\overrightarrow{r}=x\hat{i}+y\hat{j}$

$=\frac{V_o}{\sqrt{2}g}\hat{i}+\frac{V_o^2}{2g}\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]\hat{j}$

The angular momentum about point of start is

$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{V})$

$=m[\frac{V_o}{\sqrt{2}g}\hat{i}+\frac{V_o^2}{2g}\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]\hat{j}]\times [\frac{V_o}{\sqrt{2}}\hat{i}+\frac{V_o}{\sqrt{2}}(1-\sqrt{2})\hat{j}]$

$\Rightarrow \overrightarrow{L}=(\frac{m{V}_o^2}{2g}-\frac{2m{V}_o^2}{2\sqrt{2}g})\hat{k}-\frac{m{V}_o^2}{2g}\hat{k}+\frac{m{V}_o^3}{2\sqrt{2}g}\hat{k}$

$\Rightarrow \overrightarrow{L}=-\frac{m{V}_o^3}{2\sqrt{2}g}\hat{k}$

Hence magnitude is $\frac{m{V}_o^3}{2\sqrt{2}g}$