Question bank/JEE Advanced/Physics

A particle of mass $1 \mathrm{~kg}$ is subjected to a force which depends on the position as $\vec{F}=$ $-k(x \hat{i}+y \hat{j}) k g \mathrm{ks}^{-2}$ with $k=1 \mathrm{~kg} \mathrm{~s}^{-2}$. 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 $z=0.5 \mathrm{~m}$, what is the value of $\left(x v_{y}-y v_{x}\right)$ in $m^{2} s^{-1}$?

Source: JEEBench (IIT Delhi) / JEE Adv 2022 Paper 2