A particle moving in the $x y$ plane experiences a velocity dependent force $\overrightarrow{\mathrm{F}}=k\left(v_{y} \hat{i}+v_{x} \hat{j}\right)$, where $v_{x}$ and $v_{y}$ are $x$
and $y$ components of its velocity $\vec{v}$. if $\vec{a}$ is the acceleration of the particle, then which of the following statements is true for the particle?
Correct Option: 1
(1) Given, $\vec{F}=k\left(v_{y} \hat{i}+v_{x} \hat{j}\right)$
$\therefore F_{x}=k v_{y} \hat{i}, F_{y}=k v_{x} \hat{j}$
$\frac{m d v_{x}}{d t}=k v_{y} \Rightarrow \frac{d v_{x}}{d t}=\frac{k}{m} v_{y}$
Similarly, $\frac{d v_{y}}{d t}=\frac{k}{m} v_{x}$
$\frac{d v_{y}}{d v_{x}}=\frac{v_{x}}{v_{y}} \Rightarrow \int v_{y} d v_{y}=\int v_{x} d v_{x}$
$v_{y}^{2}=v_{x}^{2}+C$
$v_{y}^{2}-v_{x}^{2}=\mathrm{constant}$
$\vec{v} \times \vec{a}=\left(v_{x} \hat{i}+v_{y} \hat{j}\right) \times \frac{k}{m}\left(v_{y} \hat{i}+v_{x} \hat{j}\right)$
$=\left(v_{x}^{2} \hat{k}-v_{y}^{2} \hat{k}\right) \frac{k}{m}=\left(v_{x}^{2}-v_{y}^{2}\right) \frac{k}{m} \hat{k}=$ constant
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