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A particle of mass $1 kg$ is subjected to a force which depends on the position as $\vec{F}=-k(x \hat{i}+y \hat{j}) kgms ^{-2}$ 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. . . . . $m^2 s^{-1}$
$3$
$4$
$5$
$6$
Solution
Torque about origin is zero
So angular momentum about origin remains conserved.
$\begin{array}{l}\left|\begin{array}{ccc}i & j & k \\ \frac{1}{\sqrt{2}} & \sqrt{2} & 0 \\ -\sqrt{2} & \sqrt{2} & \frac{2}{\pi}\end{array}\right|=\left|\begin{array}{ccc}i & j & k \\ x & y & 0.5 \\ v_x & v_y & \frac{2}{\pi}\end{array}\right| \\ \hat{i}\left[\sqrt{2} \times \frac{2}{\pi}\right]-\hat{j}\left[\frac{\sqrt{2}}{\pi}\right]+\hat{k}[1+2]=i\left[\frac{y \times 2}{\pi}-0.5 v_y\right]-\hat{j}\left[\frac{x \times 2}{\pi}-0.5 v_x\right]+k\left[x v_y-y v_x\right] \\ x v_y-y v_x=3\end{array}$
