A particle moves in the x - y plane under the action of a force F such that the value of its linear momentum P at any instant is - p -2cos t - The angle - between F and p is sinA- 45-B- 30-C- 90-D- 60-

A particle moves in the x-y plane under the action of a force F such that the value of its linear momentum $(\to P)$ at any time 't' is $P_{x} = 2 c o s t, P_{y} = 2 s i n t$ .The angle $θ$ between $\to F$ and $\to P$ at a given time t. will be

Q. A particle moves in the x-y plane under the action of a force

\to F

such that the value of its linear momentum

\to P

at any time fis

P_{x} = 2 c o t P_{y} = 2 s i n t

The angle

θ

between

\to F

and

\to P

at a given time t will be:

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A particle moves in the x−y plane under the action of a force →F such that the value of its linear momentum →p at any instant is →p=2(cost^i+sint^j). The angle θ between →F and →p is

The correct option is D 90∘→F=d→pdt=2(−sint^i+cost^j) Dot product of →F and →p is →F.→p=4(−sint^i+cost^j).(cost^i+sint^j)=0 ∴θ=90∘

A particle moves in the $x - y$ plane under the action of a force $\to F$ such that the value of its linear momentum $\to p$ at any instant is $\to p = 2 (cos t^i + sin t^j)$ . The angle $θ$ between $\to F$ and $\to p$ is

The correct option is D $90^{\circ}$
$\to F = \frac{d \to p}{d t} = 2 (- sin t^i + cos t^j)$
Dot product of $\to F$ and $\to p$ is
$\to F . \to p = 4 (- sin t^i + cos t^j) . (cos t^i + sin t^j) = 0$
$∴ θ = 90^{\circ}$