A particle moves in the x-y plane under the action of a force $\overrightarrow{\mathrm{F}}$ such that the value of its linear momentum $\overrightarrow{\mathrm{P}} \mathrm{at} \mathrm{any} \mathrm{time} \mathrm{t} \mathrm{is} {\mathrm{P}}_{\mathrm{x}} = 2 \cos \mathrm{t},$

${\mathrm{P}}_{\mathrm{y}} = 2 \sin \mathrm{t}$. The angle $\theta$ between $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{P}}$ at a given time t will be

$\mathrm{P} = \sqrt{{{\mathrm{P}}_{\mathrm{x}}}^2+{{\mathrm{P}}_{\mathrm{y}}}^2} = \sqrt{{(2 \cos \mathrm{t})}^2+{(2 \sin \mathrm{t})}^2} = 2$

If m be the mass of the body, then

Kinetic energy = $\frac{{\mathrm{P}}^2}{2\mathrm{m}} = \frac{{\left(2\right)}^2}{2\mathrm{m}} = \frac{2 }{\mathrm{m}}$

Since kinetic energy does not change with time, both work done and power are zero.

Now Power = Fv $\cos \mathrm{\theta } = 0$

As  $\mathrm{F} \ne 0, \mathrm{v} \ne 0 \therefore \cos \mathrm{\theta } \mathrm{or} \mathrm{\theta } = {90}^0$

As direction of $\overrightarrow{\mathrm{P}} \mathrm{is} \mathrm{same} \mathrm{that} \overrightarrow{\mathrm{v}} (\therefore \overrightarrow{\mathrm{P}} = \mathrm{m}\overrightarrow{\mathrm{v}}), \mathrm{hence} \mathrm{angle} \mathrm{between} \overrightarrow{\mathrm{F}} \mathrm{and} \overrightarrow{\mathrm{P}} \mathrm{is} \mathrm{equal} \mathrm{to} {90}^0$