Question

A particle moves in the $x y$ -plane under the action of a force $F$ such that the components of its linear momentum $p$ at any time $t$ are $p_{x}=2 \cos t, p_{y}=\sin t .$ The angle between $F$ and $p$ at time $t$ is

• A. $90^{\circ}$
• B. $0^{\circ}$
• C. $180^{\circ}$
• D. $30^{\circ}$

Answer 1

Answer: (A)

$p_{x}=2 \cos t, p_{y}=2 \sin t$
$\therefore p =p_{x} i +p_{y} j$
$=2 \cos t\, i +2 \sin t\, j$
Force $=$ rate of change of momentum
$=\frac{d p }{d t}$
$\Rightarrow F =-2 \sin t\, i +2 \cos t\, j$
$\therefore \cos \theta=\frac{ F \cdot p }{| F | \cdot| p |}$
$=\frac{(-2 \sin t\, i +2 \cos t\, j )(2 \cos t\, i +2 \sin t\, j )}{\sqrt{4 \sin ^{2} t+4 \cos ^{2} t} \sqrt{4 \cos ^{2} t+4 \sin ^{2} t}}$
$\Rightarrow \cos \theta =0$
$\Rightarrow \theta =90^{\circ}$