25. A particle moves in the x-y plane under the action of a force F such that the value of its linear momentum (overrightarrow (P)) at anytime t is P_(x)=2cost,p_(y)=2sint The angle Theta between overrightarrow (F) and overrightarrow (P) at a given time t. will be [MNR 1991;UPSEAT 2000] (a) Theta =0^circ (b) Theta =30^circ (c) Theta =90^circ (d) Theta =180^circ

To find the angle \(\Theta\) between the force \(\overrightarrow{F}\) and the linear momentum \(\overrightarrow{P}\), we need to use the relationship between the force, momentum, and acceleration. The force acting on the particle is given by Newton's second law:<br /><br />\[<br />\overrightarrow{F} = m \overrightarrow{a}<br />\]<br /><br />where \(m\) is the mass of the particle and \(\overrightarrow{a}\) is the acceleration. The acceleration can be found from the time derivative of the velocity, which in turn can be found from the time derivative of the momentum:<br /><br />\[<br />\overrightarrow{v} = \frac{d\overrightarrow{P}}{dt}<br />\]<br />\[<br />\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = \frac{d^2\overrightarrow{P}}{dt^2}<br />\]<br /><br />Given the momentum components:<br /><br />\[<br />P_x = 2 \cos t, \quad P_y = 2 \sin t<br />\]<br /><br />we find the velocity components by differentiating with respect to time:<br /><br />\[<br />v_x = \frac{dP_x}{dt} = -2 \sin t<br />\]<br />\[<br />v_y = \frac{dP_y}{dt} = 2 \cos t<br />\]<br /><br />Next, we find the acceleration components by differentiating the velocity components with respect to time:<br /><br />\[<br />a_x = \frac{dv_x}{dt} = -2 \cos t<br />\]<br />\[<br />a_y = \frac{dv_y}{dt} = -2 \sin t<br />\]<br /><br />The force components can be found using Newton's second law:<br /><br />\[<br />F_x = m a_x = -2m \cos t<br />\]<br />\[<br />F_y = m a_y = -2m \sin t<br />\]<br /><br />The angle \(\Theta\) between the force and the momentum is given by:<br /><br />\[<br />\tan \Theta = \left| \frac{F_y}{F_x} \right| = \left| \frac{-2m \sin t}{-2m \cos t} \right| = \left| \frac{\sin t}{\cos t} \right| = \left| \tan t \right|<br />\]<br /><br />Since \(\tan t = \tan(\pi - t)\), we have:<br /><br />\[<br />\tan \Theta = \tan(\pi - t)<br />\]<br /><br />This implies that \(\Theta = 90^\circ\).<br /><br />Therefore, the correct answer is:<br /><br />(c) \(\Theta = 90^\circ\)