Question

A particle of mass $1\, kg$ is subjected to a force which depends on the position as $\vec{F}=-k(x \hat{i}+y \hat{j}) kg\, ms ^{-2}$ with $k=1 \, kgs ^{-2}$. 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 $z=0.5\, m$, the value of $\left(x v_y-y v_x\right.$ ) is _______$m^2 s^{-1}$.

Answer 1

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}-0.5 v _y\right]-\hat{ j }\left[\frac{ x \times 2}{\pi}-0.5 v _x\right]+ k \left[ xv _y- yv v _x\right] \\ x v_y-y v_x=3 \\ \end{array} $