A particle moves in the x-y plane under the action of a force F- such that value of its linear momentum p at any instant is p-2cos t -sin t - The angle - between F and p is

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Standard XII

Physics

Work Done as Dot Product

A particle mo...

Question

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

60^{\circ}

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45^{\circ}

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30^{\circ}

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90^{\circ}

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Solution

The correct option is D $90^{\circ}$
Given that the particle moves under the action of force and hence Force is equal to the rate of change of momentum(from Newtons II Law)
Linear momentum $p = 2 (c o s t^i + s i n t^j)$
$F = \frac{d p}{d t} = 2 (- s i n t^i + c o s t^j)$ and
$\to F \cdot \to p = 4 (- s i n t^i + c o s t^j) \cdot (c o s t^i + s i n t^j) = 0$
$θ = 90^{\circ}$

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

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