Question

A particle moving in the xy plane experiences a velocity dependent force \(\vec F=k\left(v_y\hat i+v_x\hat j\right)\), 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?

• A. kinetic energy of particle is constant in time
• B. quantity \(\vec v\times\vec a\) is constant in time
• C. quantity \(\vec v\cdot\vec a\) is constant in time
• D. \(\vec F\) arises due to a magnetic field

Hint:

With \(\vec a=\vec F/m\): compute \(\vec v\cdot\vec a\), \(\vec v\times\vec a\) and their time derivatives. A magnetic force would be perpendicular to \(\vec v\).

Answer 1

The Correct Option is B

\(m\vec a=\vec F\Rightarrow a_x=\dfrac{k}{m}v_y,\ a_y=\dfrac{k}{m}v_x\).
Power: \(\vec F\cdot\vec v=k(v_yv_x+v_xv_y)=2kv_xv_y\), which is not zero in general, so KE changes: (1) false, and (4) false (a magnetic force is always perpendicular to \(\vec v\), i.e. does no work).
\(\vec v\cdot\vec a=\dfrac{2k}{m}v_xv_y\) is not constant (it changes with time): (3) false.
\(\vec v\times\vec a\) has only a z-component: \(v_xa_y-v_ya_x=\dfrac{k}{m}(v_x^2-v_y^2)\).
Its time derivative: \(\dfrac{d}{dt}(v_x^2-v_y^2)=2v_xa_x-2v_ya_y=\dfrac{2k}{m}(v_xv_y-v_yv_x)=0\). So \(\vec v\times\vec a\) is constant.
Answer: (2) quantity \(\vec v\times\vec a\) is constant in time