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Question: A particle moves in the x-y plane under the action of a force \(\overset{\rightarrow}{F}\) such that...

A particle moves in the x-y plane under the action of a force F→\overset{\rightarrow}{F} such that the value of its linear momentum (P→)(\overset{\rightarrow}{P}) at anytime t is Px=2cos⁡t,py=2sin⁡t.P_{x} = 2\cos t,p_{y} = 2\sin t. The angle θ\thetabetween F→\overset{\rightarrow}{F} and P→\overset{\rightarrow}{P} at a given time t. will be

A

θ=0∘\theta = 0{^\circ}

B

θ=30∘\theta = 30{^\circ}

C

θ=90∘\theta = 90{^\circ}

D

θ=180∘\theta = 180{^\circ}

Answer

θ=90∘\theta = 90{^\circ}

Explanation

Solution

Px=2cos⁡tP_{x} = 2\cos t, Py=2sin⁡tP_{y} = 2\sin t ∴P→=2cos⁡t 6mui^+2sin⁡t 6muj^\overrightarrow{P} = 2\cos t\mspace{6mu}\widehat{i} + 2\sin t\mspace{6mu}\widehat{j}

F→=dP→dt=−2sin⁡t 6mui^+2cos⁡t 6muj^\overrightarrow{F} = \frac{d\overrightarrow{P}}{dt} = - 2\sin t\mspace{6mu}\widehat{i} + 2\cos t\mspace{6mu}\widehat{j}

F→.P→=0\overrightarrow{F}.\overrightarrow{P} = 0∴θ=90∘\theta = 90{^\circ}