A particle moving in the xy plane experiences a velocity dependent force
$\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)$ , where v_x and v_y are the
x and y components of its velocity $\overrightarrow v$ . If $\overrightarrow a$ is the
acceleration of the particle, then
which of the following statements is true for the particle?

Given $\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)$ <br><br>$\Rightarrow$ m$\overrightarrow a$ = $k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)$ <br><br>$\Rightarrow$ $$\overrightarrow a = {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} \right)$$ <br><br>Also ${{d{v_x}} \over {dt}} = {k \over m}{v_x}$ <br><br>and ${{d{v_y}} \over {dt}} = {k \over m}{v_y}$ <br><br>${{d{v_x}} \over {d{v_y}}} = {{{v_y}} \over {{v_x}}}$ <br><br>$\Rightarrow$ $\int {{v_x}d{v_x}} = \int {{v_y}} d{v_y}$ <br><br>$\Rightarrow$ $v_y^2 = v_x^2 + C$ <br><br>$\Rightarrow$ $v_y^2 - v_x^2 = C$ = Constant <br><br>From Option (B), <br><br>$\overrightarrow v \times \overrightarrow a$ <br><br>= $$\left( {{v_x}\widehat i + {v_y}\widehat j} \right) \times {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} \right)$$ <br><br>= $\left( {v_x^2\widehat k - v_y^2\widehat k} \right) \times {k \over m}$ <br><br>= $\left( {v_x^2 - v_y^2} \right) \times {k \over m}\widehat k$ <br><br>= Constant