Question

A particle is projected at time t = 0 from a point P with a speed $v_0$ at an angle of ${45}^{\circ }$ to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time $t=\frac{v_0}{g}$

• A. $\frac{m{v}_0^2}{2\sqrt{2}g}(-\hat{k})$
• B. $\frac{m{v}_0^3}{2\sqrt{2}g}(-\hat{k})$
• C. $\frac{m{v}_0^2}{2\sqrt{2}g}(-\hat{k})$
• D. $\frac{m{v}_0^3}{2\sqrt{2}g}(-\hat{k})$

Answer 1

The correct option is B

$\frac{m{v}_0^3}{2\sqrt{2}g}(-\hat{k})$

Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upward direction as shown in figure. For horizontal motion during the time 0 to t

$v_x=v_{0}cos{45}^{\circ }=\frac{v_0}{\sqrt{2}}$

and $x=v_{x}t=\frac{v_0}{\sqrt{2}}.\frac{v_0}{g}=\frac{v_0^2}{\sqrt{2}g}.$
For vertical motion,
$v_y=v_{0}sin{45}^{\circ }-gt=\frac{v_0}{\sqrt{2}}-v_0=\frac{(1-\sqrt{2})}{\sqrt{2}}{v}_0$
and $y=\left(v_{0}sin{45}^{\circ }\right)t-\frac{1}{2}g{t}^2$
$=\frac{v_0^2}{\sqrt{2}g}-\frac{v_0^2}{2g}=\frac{v_0^2}{2g}(\sqrt{2}-1)$

The angular momentum of the particle at time t about the origin is

$L=\overrightarrow{r}x\overrightarrow{p}=m\overrightarrow{r}x\overrightarrow{v}=m(\overrightarrow{i}x+\overrightarrow{j}y)x(\overrightarrow{i}{v}_x+\overrightarrow{j}{v}_y)$
$=m(\overrightarrow{k}x{v}_y-\overrightarrow{k}y{v}_x)$
$=m\overrightarrow{k}[\left(\frac{v_0^2}{\sqrt{2}g}\right)\frac{v_0}{\sqrt{2}}(1-\sqrt{2})-\frac{v_0^2}{2g}(\sqrt{2}-1)\frac{v_0}{\sqrt{2}}]$
$=-\overrightarrow{k}\frac{m{v}_0^3}{2\sqrt{2}g}$
Thus, the angular momentum of the particle is $\frac{m{v}_0^3}{2\sqrt{2}g}$ in the negative Z-

direction, i.e., perpendicular to the plane of motion, going into the plane.