Question

A particle moves in the $XY$-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 momentum is :-

• A. $0^{\circ}$
• B. $30^{\circ}$
• C. $45^{\circ}$
• D. $90^{\circ}$

Answer 1

Answer: (D)

$\vec{ p }( t )= A (\hat{ i } \cos kt -\hat{ j } \sin kt )$
$\vec{ F }=\frac{ d }{ dt }[\vec{ p }( t )]= Ak (-\hat{ i } \sin kt -\hat{ j } \cos kt )$
$\vec{ F } \cdot \vec{ p }= A ^{2} k (-\cos kt \sin kt +\sin kt \cos kt )=0$