Question

A particle is projected at time $t=0$ from a point $P$ on the ground with a speed $V_0$ at an angle of ${45}^o$ 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 $\frac{m{v}_0^3}{2\sqrt{2}g}$

$u_x=v_0\cos \theta =\frac{v_0}{\sqrt{2}}{u}_y=v_0\sin \theta -gt=\frac{v_0}{\sqrt{2}}-\frac{v_0}{g}g=v_0\left[\frac{1-\sqrt{2}}{\sqrt{2}}\right]\overline{P}=\frac{m{v}_0}{\sqrt{2}}\hat{i}+m{v}_0\left[\frac{1-\sqrt{2}}{\sqrt{2}}\right]\hat{j}\overline{r}=x\hat{i}+y\hat{j}x=uxt=\frac{v_0}{\sqrt{2}}.\frac{v_0}{g}=\frac{v_0^2}{\sqrt{2}}\frac{v_0}{g}=\frac{v_0^2}{\sqrt{2}g}$

$y=v_0\sin \theta t-\frac{1}{2}g{t}^2=\frac{v_0}{\sqrt{2}}\frac{v_0}{g}-\frac{1}{2}g\frac{v_0^2}{g^2}=\frac{v_0^2}{g^2}[\frac{1}{\sqrt{2}}-\frac{1}{2}]=\frac{v_0^2}{g^2}\left[\frac{2-\sqrt{2}}{2\sqrt{2}}\right]L=\overline{r}\times \overline{P}$

$=[\frac{v_0^2}{\sqrt{2}g}+\frac{v_0^2}{g^2}\left[\frac{2-\sqrt{2}}{2\sqrt{2}}\right]\hat{j}]\times [\frac{m{v}_0}{\sqrt{2}}\hat{i}+m{v}_0\left[\frac{1-\sqrt{2}}{\sqrt{2}}\right]\hat{j}]=-\frac{m{v}_0^3}{2\sqrt{2}g}$

$\left|\overline{L}\right|=\frac{m{v}_0^3}{2\sqrt{2}g}$