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A particle moves in the XY plane under the action of a force F→ such that the value of its linear momentum (P→) at any time 't' is Px = 2 cos t, Py = 2 sin t . The angle ' θ ' between F→ and P→ at a given time 't' will be

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a

θ = 00

b

θ = 300

c

θ = 900

d

θ = 1800

answer is C.

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Detailed Solution

p→=2costi^+2sintj^F→=dp→dt=-2sinti^+2costj^p→·F→=0⇒p→⊥ to F→

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