A particle moving in the $xy$-plane experiences a velocity dependent force $\vec F = k(v_y\hat i + v_x\hat j)$, where $v_x$ and $v_y$ are the $x$ 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?
Answer: (A) Quantity $\vec v\times\vec a$ is constant in time
With $\vec a = \vec F/m$: $ma_x = kv_y$ and $ma_y = kv_x$.
**Kinetic energy:** $\vec F\cdot\vec v = 2kv_xv_y \neq 0$ in general, so KE changes; the force is not magnetic (a magnetic force is always perpendicular to $\vec v$).
**$\vec v\times\vec a$:**
$$\vec v\times\vec a = (v_xa_y - v_ya_x)\hat k = \frac km\left(v_x^2 - v_y^2\right)\hat k$$
$$\frac{d}{dt}\left(v_x^2 - v_y^2\right) = 2v_x\frac{kv_y}{m} - 2v_y\frac{kv_x}{m} = 0$$
So $\vec v\times\vec a$ is constant in time. ($\vec v\cdot\vec a = \frac{2k}{m}v_xv_y$ changes.)
Solution by Sreeraj P, M.Sc Physics