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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Given that \(\hat{\mathbf{p}} = p_x \hat{i} + p_y \hat{j} = 2 \cos t \hat{i} + 2 \sin t \hat{j}\)
∴ ∴ \(\boxed{\mathbf{F}} = \frac{d\mathbf{p}^{\Delta}}{dt} = -2 \sin t \hat{\imath} + 2 \cos t \hat{\jmath}\)
Now, \(|\Delta| \quad |\times| \\ \mathbf{F} \cdot p = 0\) i.e. angle between \(|\Delta| F \text{ and } |\Delta| p\) is 90°.
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