Q. A particle moves in the X-Y plane under the influence of a force such that its linear momentum is $\mathbf{p}(t) = A\lbrack\widehat{\dot{\mathbf{i}}}\cos(kt) - \widehat{\dot{\mathbf{j}}}\sin{(kt)\rbrack}$ where A and $k$ are constant. The angle between the force and the momentum is
A) $0{^\circ}$
B) $30{^\circ}$
C) $45{^\circ}$
D) $90{^\circ}$
Explanation
(d) $Force\ \mathbf{F =}\frac{\mathbf{dp}}{dt} = - k\ A\sin{(kt)\ \widehat{\dot{\mathbf{i}}}} - k\ A\cos{(kt)\ \widehat{\dot{\mathbf{j}}}}$
$\mathbf{p =}A\cos{(kt)\ \widehat{\dot{\mathbf{i}}}} - A\sin{(kt)\ \widehat{\dot{\mathbf{j}}}}$
$Since,\ \ \mathbf{F \bullet p =}0$
$\therefore\ $Angle between F and p should be $90{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.