Question

A particle is projected at time $t=0$ from a point P on the ground with a speed $u$ at an angle ${45}^o$ to the horizontal. The magnitude of the angular momentum of the particle about the point P at the time $t=\frac{u}{g\sqrt{2}}$ is:

• A. $\frac{m{u}^3}{\sqrt{2}g}$
• B. $\frac{m{u}^3}{2g}$
• C. $\frac{m{u}^3}{4\sqrt{2g}}$
• D. $Zero$

Answer 1

The correct option is C $\frac{m{u}^3}{4\sqrt{2g}}$

Time of flight $T=\frac{2u\sin \theta }{g}=\frac{2u\sin {45}^0}{g}=\frac{2u\frac{1}{\sqrt{2}}}{g}=2\times \left(\frac{u}{g\sqrt{2}}\right)$
At time $t=\frac{u}{g\sqrt{2}}$ the particle will be at highest point on the trajectory since $t=\frac{T}{2}$
At the highest point on the trajectory, velocity is horizontal since vertical component of velocity is zero there.
$u_x=u\cos {45}^0=\frac{u}{\sqrt{2}}$
Angular momentum $\overrightarrow{L}$ $=\overrightarrow{r}$ x$\overrightarrow{p}$
$\overrightarrow{v}$ and $\overrightarrow{r}$ are perpendicular to each other at the highest point.
$\overrightarrow{r}=\frac{u^{2}si{n}^2\theta }{2g}\hat{j}$ since the particle is at the highest point
$\overrightarrow{p}=m{u}_x=m\frac{u}{\sqrt{2}}\hat{i}$
$\therefore \overrightarrow{L}$ $=\frac{u^{2}si{n}^2{45}^0}{2g}\hat{j}$ x $m\frac{u}{\sqrt{2}}\hat{i}=\frac{m{u}^3}{4\sqrt{2}g}(-\hat{k})$
$\therefore \left|\overrightarrow{L}\right|=\frac{m{u}^3}{4\sqrt{2}g}$