A particle moves in xy-plane under the influence of a force such that its linear momentum is p→t=Ai^coskt−j^sinkt , where A and k are constants. The angle between the force and the momentum is

F→=dpdt=ddtAi^coskt−j^sinkt= − kAi^sinkt+j^cosktF→⋅p→= − kAi^sinkt+j^cosktAi^coskt−Aj^sinkt=0Angle between F→⋅p→ is 900

A particle moves in xy-plane under the influence of a force such that its linear momentum is p→t=Ai^coskt−j^sinkt , where A and k are constants. The angle between the force and the momentum is

F→=dpdt=ddtAi^coskt−j^sinkt= − kAi^sinkt+j^cosktF→⋅p→= − kAi^sinkt+j^cosktAi^coskt−Aj^sinkt=0Angle between F→⋅p→ is 900

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A particle moves in xy-plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ , where $A$ and $k$ are constants. The angle between the force and the momentum is

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$0^{0}$

$30^{0}$

$45^{0}$

$90^{0}$

answer is D.

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Detailed Solution

$\vec{F} = \frac{d p}{d t} = \frac{d}{d t} [A (\hat{i} \cos k t - \hat{j} \sin k t)]$

$= - k A (\hat{i} \sin k t + \hat{j} \cos k t)$

$\vec{F} \cdot \vec{p} = - k A (\hat{i} \sin k t + \hat{j} \cos k t) (A \hat{i} \cos k t - A \hat{j} \sin k t)$

$= 0$

Angle between $\vec{F} \cdot \vec{p}$ is $90^{0}$

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A particle moves in xy-plane under the influence of a force such that its linear momentum is p→t=Ai^coskt−j^sinkt , where A and k are constants. The angle between the force and the momentum is

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A particle moves in xy-plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ , where $A$ and $k$ are constants. The angle between the force and the momentum is