A particle moves in the $$X - Y$$  plane under the influence of a force such that its linear momentum is $$\vec p(t) = A[\hat i\cos (kt) - \hat j\sin (kt)],$$       where $$A$$ and $$k$$ are constants. The angle between the force and the momentum is

Solution :
$$\eqalign{ & \vec p\left( t \right) = A\left[ {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right] \cr & \vec F = \frac{{d\vec p}}{{dt}} = Ak\left[ { - \hat i\sin \left( {kt} \right) - \hat j\cos \left( {kt} \right)} \right] \cr & {\text{Here,}}\,\vec F.\vec P = 0\,\,{\text{But}}\,\,\vec F.\vec P = Fp\cos \theta \cr & \therefore \cos \theta = 0 \Rightarrow \theta = {90^ \circ } \cr} $$