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JEE Main Physics question

PhysicsMotion in a PlaneJEE Main

A particle moves in the x-y plane under the action of a force F, such that its linear momentum p at a given time t is pₓ = 2cos t, = 2sin t. Then the angle θ between F and p at that time t is

Answer: (D)

Solution

F = dp/dt = (−2sin t, 2cos t). p·F = (2cos t)(−2sin t) + (2sin t)(2cos t) = 0, so F is always perpendicular to p; θ = 90°.

Question ID VK-152873