Question

A particle of mass $m$ is projected at time $t=0$ 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 $t=\frac{v_0}{g}$?

• 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 $\frac{m{v}_0^3}{2\sqrt{2}g}$

Assume upward motion as positive.


At time $t=\frac{v_0}{g}$:
Horizontal component of speed $v_x=u_x=\frac{v_0}{\sqrt{2}}$
Vertical component of speed $\left(v_y\right)$
from $(v=u+at)$
$v_y=u_y+a_{y}t$ (Here $a_y=-g$)
$v_y=v_0\sin {45}^{\circ }-g\times \frac{v_0}{g}=v_0(\frac{1}{\sqrt{2}}-1)$
Therefore, velocity at $t=\frac{v_0}{g}$
$\overrightarrow{v}=\frac{v_0}{\sqrt{2}}\hat{i}+v_0(\frac{1}{\sqrt{2}}-1)\hat{j}$ ... (1)

Distance travelled in $x-$ direction $S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$ where $(a_x=0)$
$S_x=\frac{v_0}{\sqrt{2}}\times \frac{v_0}{g}=\frac{v_0^2}{\sqrt{2}g}$
Distance travelled in $y-$ direction in $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$ where $(a_y=-g)$
$S_y=v_0\sin {45}^{\circ }\times \frac{v_0}{g}-\frac{1}{2}\times g\times {\left(\frac{v_0}{g}\right)}^2$
$S_y=\frac{v_0^2}{\sqrt{2}g}-\frac{v_0^2}{2g}=\frac{v_0^2}{g}\left[\frac{2-\sqrt{2}}{2\sqrt{2}}\right]$
Hence, position vector of particle at time $t=\frac{v_0}{g}$ w.r.t. point $P$ will be,
$\overrightarrow{r}=\frac{v_0^2}{\sqrt{2}g}\hat{i}+\frac{v_0^2}{g}\left[\frac{2-\sqrt{2}}{2\sqrt{2}}\right]\hat{j}$ ... (2)

As we know, $\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
From eqn. (1) and (2)
$\overrightarrow{L}=m[(\frac{v_0^2}{\sqrt{2}g}\hat{i}+\frac{v_0^2}{g}\left[\frac{2-\sqrt{2}}{2\sqrt{2}}\right]\hat{j})\times (\frac{v_0}{\sqrt{2}}\hat{i}+v_0\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)\hat{j})]$
$=m\left[\frac{v_0^3(1-\sqrt{2})}{2g}\right(\hat{i}\times \hat{j})+\frac{v_0^3(2-\sqrt{2})}{4g}(\hat{j}\times \hat{i}\left)\right]$
$[\because \hat{i}\times \hat{i}=0=\hat{j}\times \hat{j}\&\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{i}=(-\hat{k}\left)\right]$
$\overrightarrow{L}=m\left[\frac{v_0^3}{4g}[2-2\sqrt{2}-2+\sqrt{2}]\hat{k}\right]=\frac{m{v}_0^3\sqrt{2}}{4g}(-\hat{k})$
(-ve sign means -ve $z$ direction)
$\therefore \overrightarrow{L}=\frac{m{v}_0^3}{2\sqrt{2}g}(-\hat{k})$