Question

A particle of mass $1kg$ is subjected to a force which depends on the position as $\vec{F}=-k(x\hat{i}+y\hat{j})kgm{s}^{-2}$ with $k=1kg{s}^{-2}$. At time $t=0$, the particle's position $\vec{r}=\left(\frac{1}{\sqrt{2}}\hat{i}+\sqrt{2}\hat{j}\right)m$ and its velocity $\vec{v}=\left(-\sqrt{2}\hat{i}+\sqrt{2}\hat{j}+\frac{2}{\pi }\hat{k}\right)m{s}^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5m$, the value of $(x{v}_y-y{v}_x$ ) is _______$m^2{s}^{-1}$.

Answer 1

Answer: 3

Torque about origin is zero So angular momentum about origin remains conserved.
$\begin{array}{l}\left|\begin{array}{ccc}i& j& k\\ \frac{1}{\sqrt{2}}& \sqrt{2}& 0\\ -\sqrt{2}& \sqrt{2}& \frac{2}{\pi }\end{array}\right|=\left|\begin{array}{ccc}i& j& k\\ x& y& 0.5\\ v_x& v_y& \frac{2}{\pi }\end{array}\right|\\ \hat{i}\left[\sqrt{2}\times \frac{2}{\pi }\right]-\hat{j}\left[\frac{\sqrt{2}}{\pi }\right]+\hat{k}[1+2]=i\left[\frac{y\times 2}{\pi }-0.5v_y\right]-\hat{j}\left[\frac{x\times 2}{\pi }-0.5v_x\right]+k\left[x{v}_y-yv{v}_x\right]\\ x{v}_y-y{v}_x=3\end{array}$