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

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$180^{\circ}$

$30^{\circ}$

$0^{\circ}$

$90^{\circ}$

answer is A.

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Detailed Solution

$Given that \vec{p} = p_{x} \hat{i} + p_{y} \hat{j} = 2 \cos t \hat{i} + 2 \sin t \hat{j}$

$∴ \vec{F} = \frac{d \vec{p}}{d t} = - 2 \sin t \hat{i} + 2 \cos t \hat{j}$

$Now, \vec{F} \cdot \vec{p} = 0 i.e. angle between \vec{F} and \vec{p} is 90^{\circ}$

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