A particle moves in the X-Y plane under the influence of a force such that its linear momentum is $p(t) =A[\widehat{i}cos\, (kt) -\widehat{j}sin\, (kt)]$ where, A and k are constants. The angle between the force and the momentum is
- $0^\circ$
- $30^\circ$
- $45^\circ$
- $90^\circ$
The Correct Option is D
Solution and Explanation
$ \, \, \, \, \, \, \, p=A\, cos\, (kt) \widehat{i} - A\, sin\, (kt)\widehat{j}$
Since, $\hspace20mm F.p=0$
$\therefore$ Angle between F and p should be $90^\circ.$
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